Samacheer Kalvi Class 11 Maths Solutions Chapter 9 Limits and Continuity Exercise 9.4

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.4

Evaluate the following limits:

 

Question 1. \( \lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^{7x} \)
Answer: We know that \( \lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^x = e \).
So, we can rewrite the given limit:
\( \lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^{7x} = \left[ \lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^x \right]^7 \)
\( \implies = e^7 \)
This limit is a fundamental definition of the mathematical constant 'e', which is important in many areas like compound interest and natural growth.
In simple words: When you see \( \left(1 + \frac{1}{x}\right)^x \) and \( x \) goes to infinity, it becomes \( e \). If there's an extra power like \( 7x \), it means \( e \) will be raised to that power, making the answer \( e^7 \).

🎯 Exam Tip: Remember the two basic forms for 'e' limits: \( \lim_{x\to\infty} (1+1/x)^x = e \) and \( \lim_{x\to0} (1+x)^{1/x} = e \). Adapt the given limit to one of these forms.

 

Question 2. \( \lim_{x\to0} (1 + x)^{1/(3x)} \)
Answer: We know that \( \lim_{x\to0} (1 + x)^{1/x} = e \).
We can rewrite the given limit:
\( \lim_{x\to0} (1 + x)^{\frac{1}{3x}} = \lim_{x\to0} \left[ (1 + x)^{\frac{1}{x}} \right]^{\frac{1}{3}} \)
\( \implies = e^{\frac{1}{3}} \)
This form of the limit for 'e' is particularly useful when dealing with very small changes or instantaneous rates of change.
In simple words: When \( x \) gets very close to zero, the expression \( (1+x) \) raised to the power of \( 1/x \) becomes \( e \). If the power is \( 1/(3x) \), it is like \( e \) raised to the power of \( 1/3 \).

🎯 Exam Tip: Look for the standard definition of 'e' in limits. If the exponent is a multiple or fraction of the standard form, simply apply that multiple to 'e' as an exponent.

 

Question 3. \( \lim_{x\to\infty} \left(1 + \frac{k}{x}\right)^m \)
Answer: Let the given limit be \( A \).
\( A = \lim_{x\to\infty} \left(1 + \frac{k}{x}\right)^m \)
As \( x \to \infty \), the term \( \frac{k}{x} \to 0 \).
So, the expression becomes \( (1 + 0)^m \).
\( \implies A = 1^m \)
\( \implies A = 1 \)
This is a direct limit evaluation; understanding what happens to terms like \( k/x \) as \( x \) goes to infinity is key here.
In simple words: When \( x \) gets very large, the fraction \( k/x \) becomes almost zero. So, the expression inside the bracket is nearly \( (1 + 0) \), which is \( 1 \). When \( 1 \) is raised to any power \( m \), the answer is always \( 1 \).

🎯 Exam Tip: For simple limits, evaluate terms individually first. If a part approaches zero or a constant, substitute those values before applying complex limit rules.

 

Question 4. \( \lim_{x\to\infty} \left( \frac{2x^2 + 3}{2x^2 + 5} \right)^{8x^2 + 3} \)
Answer: The given limit is of the indeterminate form \( 1^\infty \). We use the property \( \lim_{x\to a} [f(x)]^{g(x)} = e^{\lim_{x\to a} [f(x)-1]g(x)} \) when \( f(x) \to 1 \) and \( g(x) \to \infty \).
Here, \( f(x) = \frac{2x^2 + 3}{2x^2 + 5} \) and \( g(x) = 8x^2 + 3 \).
First, calculate \( f(x) - 1 \):
\( f(x) - 1 = \frac{2x^2 + 3}{2x^2 + 5} - 1 = \frac{2x^2 + 3 - (2x^2 + 5)}{2x^2 + 5} = \frac{-2}{2x^2 + 5} \)
Now, calculate \( \lim_{x\to\infty} [f(x) - 1]g(x) \):
\( \lim_{x\to\infty} \left( \frac{-2}{2x^2 + 5} \right) (8x^2 + 3) \)
\( = \lim_{x\to\infty} \frac{-2(8x^2 + 3)}{2x^2 + 5} \)
To evaluate this limit, divide the numerator and denominator by the highest power of \( x \), which is \( x^2 \):
\( = \lim_{x\to\infty} \frac{-2(8 + \frac{3}{x^2})}{2 + \frac{5}{x^2}} \)
As \( x \to \infty \), \( \frac{3}{x^2} \to 0 \) and \( \frac{5}{x^2} \to 0 \).
\( = \frac{-2(8 + 0)}{2 + 0} = \frac{-16}{2} = -8 \)
Therefore, the original limit is \( e^{-8} \).
Limits of the form \( 1^\infty \) are often encountered in growth and decay problems, where a small change over time accumulates significantly.
In simple words: This problem is of a special type called \( 1 \) to the power of \( \infty \). We use a formula that changes it into \( e \) raised to a new power. This new power is found by taking the base minus \( 1 \) and multiplying it by the original exponent, then finding the limit of that result. After calculations, this new power comes out to be \( -8 \), so the final answer is \( e^{-8} \).

🎯 Exam Tip: When faced with a \( 1^\infty \) indeterminate form, convert it to \( e^{\lim (f(x)-1)g(x)} \). This simplifies the problem significantly.

 

Question 5. \( \lim_{x\to\infty} \left(1 + \frac{3}{x}\right)^{x+2} \)
Answer: We know that \( \lim_{x\to\infty} \left(1 + \frac{k}{x}\right)^x = e^k \).
We can split the given limit:
\( \lim_{x\to\infty} \left(1 + \frac{3}{x}\right)^{x+2} = \lim_{x\to\infty} \left(1 + \frac{3}{x}\right)^x \cdot \lim_{x\to\infty} \left(1 + \frac{3}{x}\right)^2 \)
For the first part, using the standard limit with \( k=3 \):
\( \lim_{x\to\infty} \left(1 + \frac{3}{x}\right)^x = e^3 \)
For the second part, as \( x \to \infty \), \( \frac{3}{x} \to 0 \):
\( \lim_{x\to\infty} \left(1 + \frac{3}{x}\right)^2 = (1 + 0)^2 = 1^2 = 1 \)
Now, multiply the results:
\( = e^3 \cdot 1 \)
\( = e^3 \)
Recognizing that the exponent can be separated helps break down complex limits into simpler, known forms.
In simple words: We can break the power \( x+2 \) into two parts: \( x \) and \( 2 \). The part \( \left(1 + \frac{3}{x}\right)^x \) becomes \( e^3 \). The other part \( \left(1 + \frac{3}{x}\right)^2 \) becomes \( (1+0)^2 \) which is \( 1 \). So, the final answer is \( e^3 \times 1 \), which is \( e^3 \).

🎯 Exam Tip: When the exponent is a sum, split the limit into a product of two limits. One part usually uses the 'e' definition, and the other part simplifies by direct substitution.

 

Question 6. \( \lim_{x\to0} \frac{\sin^3 \left(\frac{x}{2}\right)}{x^3} \)
Answer: We know that \( \lim_{\theta\to0} \frac{\sin \theta}{\theta} = 1 \).
To use this property, we need \( \frac{x}{2} \) in the denominator for \( \sin \left(\frac{x}{2}\right) \).
\( \lim_{x\to0} \frac{\sin^3 \left(\frac{x}{2}\right)}{x^3} = \lim_{x\to0} \frac{\sin^3 \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^3 \cdot 2^3} \)
\( = \lim_{x\to0} \frac{1}{2^3} \cdot \left[ \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \right]^3 \)
As \( x \to 0 \), \( \frac{x}{2} \to 0 \). So, \( \lim_{x\to0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} = 1 \).
\( = \frac{1}{8} \cdot (1)^3 \)
\( = \frac{1}{8} \)
This limit is a foundational result for understanding the behavior of trigonometric functions near zero, especially in calculus.
In simple words: We want to make the expression look like \( \sin(\theta)/\theta \) which becomes \( 1 \) when \( \theta \) is very small. Since we have \( \sin^3(x/2) \) and \( x^3 \), we need \( (x/2)^3 \) in the denominator. We adjust the fraction by multiplying and dividing by \( 2^3 \), which leaves us with \( 1/8 \) after the \( \sin(\theta)/\theta \) part turns into \( 1 \).

🎯 Exam Tip: For limits involving \( \sin^n(\text{angle})/\text{variable}^n \), adjust the denominator to match \( (\text{angle})^n \) and compensate with appropriate constants.

 

Question 7. \( \lim_{x\to0} \frac{\sin(ax)}{\sin(bx)} \)
Answer: We know that \( \lim_{\theta\to0} \frac{\sin \theta}{\theta} = 1 \).
To use this property, we can rewrite the expression:
\( \lim_{x\to0} \frac{\sin(ax)}{\sin(bx)} = \lim_{x\to0} \frac{\frac{\sin(ax)}{ax} \cdot ax}{\frac{\sin(bx)}{bx} \cdot bx} \)
\( = \lim_{x\to0} \frac{\frac{\sin(ax)}{ax}}{\frac{\sin(bx)}{bx}} \cdot \frac{ax}{bx} \)
As \( x \to 0 \), \( ax \to 0 \) and \( bx \to 0 \).
So, \( \lim_{x\to0} \frac{\sin(ax)}{ax} = 1 \) and \( \lim_{x\to0} \frac{\sin(bx)}{bx} = 1 \).
\( = \frac{1}{1} \cdot \lim_{x\to0} \frac{ax}{bx} \)
\( = 1 \cdot \frac{a}{b} \)
\( = \frac{a}{b} \)
This property is extremely useful for quickly evaluating limits involving ratios of sine functions, especially when their arguments approach zero.
In simple words: To solve this, we want to make both the top and bottom parts look like \( \sin(\theta)/\theta \), which becomes \( 1 \). We do this by dividing \( \sin(ax) \) by \( ax \) and multiplying by \( ax \), and similarly for \( \sin(bx) \). After the \( \sin(\theta)/\theta \) parts become \( 1 \), we are left with \( ax/bx \), which simplifies to \( a/b \).

🎯 Exam Tip: For ratios of sine terms like \( \sin(ax)/\sin(bx) \) as \( x \to 0 \), the limit directly simplifies to \( a/b \).

 

Question 8. \( \lim_{x\to0} \frac{\tan(2x)}{\sin(5x)} \)
Answer: We know that \( \lim_{\theta\to0} \frac{\tan \theta}{\theta} = 1 \) and \( \lim_{\theta\to0} \frac{\sin \theta}{\theta} = 1 \).
We can rewrite the expression to use these standard limits:
\( \lim_{x\to0} \frac{\tan(2x)}{\sin(5x)} = \lim_{x\to0} \frac{\frac{\tan(2x)}{2x} \cdot 2x}{\frac{\sin(5x)}{5x} \cdot 5x} \)
\( = \lim_{x\to0} \frac{\frac{\tan(2x)}{2x}}{\frac{\sin(5x)}{5x}} \cdot \frac{2x}{5x} \)
As \( x \to 0 \), \( 2x \to 0 \) and \( 5x \to 0 \).
So, \( \lim_{x\to0} \frac{\tan(2x)}{2x} = 1 \) and \( \lim_{x\to0} \frac{\sin(5x)}{5x} = 1 \).
\( = \frac{1}{1} \cdot \lim_{x\to0} \frac{2x}{5x} \)
\( = 1 \cdot \frac{2}{5} \)
\( = \frac{2}{5} \)
This example shows how to handle combinations of tangent and sine functions in limits, which is often done by converting tangent to sine over cosine, or directly using the \( \tan \theta / \theta \) limit.
In simple words: We make the `tan(2x)` part look like `tan(\theta)/\theta` by dividing by `2x` and multiplying by `2x`. We do the same for `sin(5x)`, dividing by `5x` and multiplying by `5x`. The `tan(\theta)/\theta` and `sin(\theta)/\theta` parts become `1`. What's left is `2x/5x`, which simplifies to `2/5`.

🎯 Exam Tip: For limits involving ratios of tangent and sine, use the `tan(ax)/ax \to 1` and `sin(bx)/bx \to 1` forms. The `x` terms will often cancel out.

 

Question 9. \( \lim_{a\to0} \frac{\sin(a^n)}{(\sin a)^m} \)
Answer: We know that \( \lim_{\theta\to0} \frac{\sin \theta}{\theta} = 1 \).
We can rewrite the expression using this property:
\( \lim_{a\to0} \frac{\sin(a^n)}{(\sin a)^m} = \lim_{a\to0} \frac{\frac{\sin(a^n)}{a^n} \cdot a^n}{\left(\frac{\sin a}{a} \cdot a\right)^m} \)
\( = \lim_{a\to0} \frac{\frac{\sin(a^n)}{a^n} \cdot a^n}{\left(\frac{\sin a}{a}\right)^m \cdot a^m} \)
As \( a \to 0 \), \( a^n \to 0 \), so \( \lim_{a\to0} \frac{\sin(a^n)}{a^n} = 1 \). Also, \( \lim_{a\to0} \frac{\sin a}{a} = 1 \).
\( = \frac{1 \cdot a^n}{1^m \cdot a^m} = \lim_{a\to0} a^{n-m} \)

We need to consider three cases for the value of \( n-m \):
**Case (i) \( m = n \):**
If \( m=n \), then \( n-m = 0 \).
\( \lim_{a\to0} a^{n-m} = \lim_{a\to0} a^0 = 1 \).

**Case (ii) \( m > n \):**
If \( m > n \), then \( n-m < 0 \). Let \( n-m = -k \) where \( k > 0 \).
\( \lim_{a\to0} a^{n-m} = \lim_{a\to0} a^{-k} = \lim_{a\to0} \frac{1}{a^k} \).
As \( a \to 0 \), \( \frac{1}{a^k} \) approaches \( \infty \).

**Case (iii) \( m < n \):**
If \( m < n \), then \( n-m > 0 \). Let \( n-m = k \) where \( k > 0 \).
\( \lim_{a\to0} a^{n-m} = \lim_{a\to0} a^k \).
As \( a \to 0 \), \( a^k \) approaches \( 0 \).
This problem is a great example of how understanding exponents and fundamental trigonometric limits is crucial for solving more complex indeterminate forms.
In simple words: We adjust the expression so that we can use the rule that \( \sin(\text{angle})/\text{angle} \) becomes \( 1 \) when the angle is very small. This simplifies the expression to \( a \) raised to the power of \( (n-m) \). The final answer depends on whether \( n \) is equal to, greater than, or less than \( m \). If \( n=m \), the answer is \( 1 \). If \( m \) is bigger than \( n \), the answer goes to \( \infty \). If \( n \) is bigger than \( m \), the answer is \( 0 \).

🎯 Exam Tip: When a limit involves exponents like `n` and `m`, always consider different cases for their relationship (equal, greater than, less than) as this determines the limit's behavior.

 

Question 10. \( \lim_{x\to0} \frac{\sin(a+x) - \sin(a-x)}{x} \)
Answer: We use the trigonometric identity: \( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \).
Let \( C = a+x \) and \( D = a-x \).
Then \( C+D = (a+x) + (a-x) = 2a \), so \( \frac{C+D}{2} = a \).
And \( C-D = (a+x) - (a-x) = 2x \), so \( \frac{C-D}{2} = x \).
Substitute these into the identity:
\( \sin(a+x) - \sin(a-x) = 2 \cos(a) \sin(x) \)
Now, substitute this back into the limit:
\( \lim_{x\to0} \frac{2 \cos(a) \sin(x)}{x} \)
Since \( \cos(a) \) is a constant with respect to \( x \), we can take it out of the limit:
\( = 2 \cos(a) \lim_{x\to0} \frac{\sin(x)}{x} \)
We know that \( \lim_{x\to0} \frac{\sin(x)}{x} = 1 \).
\( = 2 \cos(a) \cdot 1 \)
\( = 2 \cos(a) \)
Knowing trigonometric sum-to-product identities can transform complex expressions into simpler forms that are easier to evaluate using standard limits.
In simple words: We use a special formula to change the top part, `sin(a+x) - sin(a-x)`, into `2 cos(a) sin(x)`. Then, the problem becomes `lim (2 cos(a) sin(x)) / x`. We know that `sin(x)/x` becomes `1` when `x` is very small. So, the whole thing simplifies to `2 cos(a)` multiplied by `1`.

🎯 Exam Tip: When you see a difference of sine or cosine functions in a limit, consider using sum-to-product or product-to-sum trigonometric identities to simplify the expression.

 

Question 11. \( \lim_{x\to0} \frac{\sqrt{x^2 + a^2} - a}{\sqrt{x^2 + b^2} - b} \)
Answer: This limit is of the indeterminate form \( \frac{0}{0} \). We can solve it by rationalizing both the numerator and the denominator.
Multiply the expression by \( \frac{\sqrt{x^2 + a^2} + a}{\sqrt{x^2 + a^2} + a} \) and \( \frac{\sqrt{x^2 + b^2} + b}{\sqrt{x^2 + b^2} + b} \):
\( \lim_{x\to0} \frac{\sqrt{x^2 + a^2} - a}{\sqrt{x^2 + b^2} - b} \cdot \frac{\sqrt{x^2 + a^2} + a}{\sqrt{x^2 + a^2} + a} \cdot \frac{\sqrt{x^2 + b^2} + b}{\sqrt{x^2 + b^2} + b} \)
Using the identity \( (A-B)(A+B) = A^2 - B^2 \):
\( = \lim_{x\to0} \frac{(x^2 + a^2 - a^2)(\sqrt{x^2 + b^2} + b)}{(x^2 + b^2 - b^2)(\sqrt{x^2 + a^2} + a)} \)
\( = \lim_{x\to0} \frac{x^2(\sqrt{x^2 + b^2} + b)}{x^2(\sqrt{x^2 + a^2} + a)} \)
Cancel out \( x^2 \) from the numerator and denominator (since \( x \ne 0 \) in the limit process):
\( = \lim_{x\to0} \frac{\sqrt{x^2 + b^2} + b}{\sqrt{x^2 + a^2} + a} \)
Now, substitute \( x=0 \):
\( = \frac{\sqrt{0^2 + b^2} + b}{\sqrt{0^2 + a^2} + a} \)
\( = \frac{\sqrt{b^2} + b}{\sqrt{a^2} + a} \)
\( = \frac{b + b}{a + a} \)
\( = \frac{2b}{2a} \)
\( = \frac{b}{a} \)
Rationalizing the numerator and denominator, or using the generalized limit form for \( (x^n - a^n)/(x-a) \), are both powerful techniques for evaluating limits involving radicals.
In simple words: This problem has square roots on both the top and bottom. To remove them, we multiply both the top and bottom by their "conjugates" (the same terms with a plus sign). This helps us get rid of the roots and cancel out `x^2`. Then, we can put `x=0` into the simplified expression, which gives us `b/a`.

🎯 Exam Tip: For limits involving square roots in both the numerator and denominator resulting in `0/0`, always rationalize both parts by multiplying by their respective conjugates.

 

Question 12. \( \lim_{x\to0} \frac{2 \arcsin x}{3x} \)
Answer: We know that \( \lim_{x\to0} \frac{\arcsin x}{x} = 1 \).
We can rewrite the given limit:
\( \lim_{x\to0} \frac{2 \arcsin x}{3x} = \frac{2}{3} \lim_{x\to0} \frac{\arcsin x}{x} \)
\( = \frac{2}{3} \cdot 1 \)
\( = \frac{2}{3} \)
The limit \( \lim (\arcsin x)/x = 1 \) is analogous to \( \lim (\sin x)/x = 1 \), highlighting the inverse relationship between the functions.
In simple words: We know that when \( x \) gets very close to zero, `arcsin(x)` divided by `x` equals `1`. In this problem, we can take the fraction `2/3` out of the limit. The remaining part, `arcsin(x)/x`, becomes `1`. So, the final answer is simply `2/3`.

🎯 Exam Tip: Remember standard inverse trigonometric limits like `lim (arcsin x)/x = 1` and `lim (arctan x)/x = 1` as `x \to 0`.

 

Question 13. \( \lim_{x\to0} \frac{1 - \cos x}{x^2} \)
Answer: We know the trigonometric identity \( 1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right) \).
Substitute this into the limit expression:
\( \lim_{x\to0} \frac{2 \sin^2 \left(\frac{x}{2}\right)}{x^2} \)
To use the standard limit \( \lim_{\theta\to0} \frac{\sin \theta}{\theta} = 1 \), we need \( \left(\frac{x}{2}\right)^2 \) in the denominator for \( \sin^2 \left(\frac{x}{2}\right) \).
\( = \lim_{x\to0} 2 \cdot \frac{\sin^2 \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2 \cdot 4} \)
\( = \lim_{x\to0} \frac{2}{4} \cdot \left( \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \right)^2 \)
\( = \frac{1}{2} \cdot \left( \lim_{x\to0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} \right)^2 \)
As \( x \to 0 \), \( \frac{x}{2} \to 0 \), so \( \lim_{x\to0} \frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}} = 1 \).
\( = \frac{1}{2} \cdot (1)^2 \)
\( = \frac{1}{2} \)
This limit is a direct consequence of the small-angle approximation \( 1 - \cos x \approx x^2/2 \) for small \( x \).
In simple words: We use the identity `1 - cos x` is equal to `2 sin^2(x/2)`. Then, we make the `sin(x/2)` part look like `sin(\theta)/\theta` which becomes `1`. We also adjust the `x^2` in the bottom. After all the adjustments, it simplifies to `2` multiplied by `1/4`, which gives us `1/2`.

🎯 Exam Tip: Always remember the identity \( 1 - \cos x = 2 \sin^2(x/2) \). This is crucial for solving limits involving \( 1 - \cos x \).

 

Question 14. \( \lim_{x\to0} \frac{\tan(2x)}{x} \)
Answer: We know that \( \lim_{\theta\to0} \frac{\tan \theta}{\theta} = 1 \).
To use this property, we need \( 2x \) in the denominator for \( \tan(2x) \).
\( \lim_{x\to0} \frac{\tan(2x)}{x} = \lim_{x\to0} \frac{\tan(2x)}{2x} \cdot 2 \)
Since \( 2 \) is a constant, we can take it out of the limit:
\( = 2 \cdot \lim_{x\to0} \frac{\tan(2x)}{2x} \)
As \( x \to 0 \), \( 2x \to 0 \). So, \( \lim_{x\to0} \frac{\tan(2x)}{2x} = 1 \).
\( = 2 \cdot 1 \)
\( = 2 \)
This limit shows that for small angles, \( \tan(2x) \) is approximately \( 2x \), which is a useful linear approximation.
In simple words: We know that `tan(angle)` divided by the `angle` itself becomes `1` when the angle is very small. Here, the angle is `2x`. So, we need `2x` in the bottom. We multiply the fraction by `2/2`. Then `tan(2x)/(2x)` becomes `1`, and we are left with the `2`.

🎯 Exam Tip: For limits of `tan(ax)/x` as `x \to 0`, the limit is simply `a` because `tan(ax)/(ax)` goes to `1` and you are left with `a`.

 

Question 15. Evaluate the following limits:
\( \lim_{x \to 0} \frac { 2^x - 3^x }{ x } \)
Answer: We know the formula: \( \lim_{x \to 0} \frac { a^x - 1 }{ x } = \log a \), where \( a > 0 \).
To use this formula, we can rewrite the expression:
\( \lim_{x \to 0} \frac { 2^x - 3^x }{ x } \)
We add and subtract 1 in the numerator:
\( \lim_{x \to 0} \frac { 2^x - 1 + 1 - 3^x }{ x } \)
Now, group the terms:
\( \lim_{x \to 0} \frac { (2^x - 1) - (3^x - 1) }{ x } \)
Separate the fraction into two parts:
\( \lim_{x \to 0} \left( \frac { 2^x - 1 }{ x } - \frac { 3^x - 1 }{ x } \right) \)
Apply the limit to each part using the formula:
\( \lim_{x \to 0} \frac { 2^x - 1 }{ x } - \lim_{x \to 0} \frac { 3^x - 1 }{ x } \)
\( = \log 2 - \log 3 \)
Using the logarithm property \( \log A - \log B = \log \left( \frac { A }{ B } \right) \):
\( = \log \left( \frac { 2 }{ 3 } \right) \)
So, the final value of the limit is \( \log \left( \frac { 2 }{ 3 } \right) \). This method helps us break down complex limits into simpler, known forms.
In simple words: We changed the top part of the fraction by adding and taking away 1. This helped us split it into two parts that fit a special limit rule. Then we just used the rule for each part and subtracted the answers.

🎯 Exam Tip: When you see \( a^x - b^x \) in a limit, a common trick is to add and subtract 1 in the numerator to convert it into the form \( (a^x - 1) - (b^x - 1) \) and apply the standard limit formula for \( (a^x - 1)/x \).

 

Question 16. Evaluate the following limits:
\( \lim_{x \to 0} \frac { 3^x - 1 }{ \sqrt{x + 1} - 1 } \)
Answer: We know the formula \( \lim_{x \to 0} \frac { a^x - 1 }{ x } = \log a \). We need to manipulate the given expression to use this.
The given limit is:
\( \lim_{x \to 0} \frac { 3^x - 1 }{ \sqrt{x + 1} - 1 } \)
To deal with the square root in the denominator, we multiply the numerator and denominator by its conjugate, \( \sqrt{x + 1} + 1 \):
\( = \lim_{x \to 0} \frac { (3^x - 1) }{ (\sqrt{x + 1} - 1) } \times \frac { (\sqrt{x + 1} + 1) }{ (\sqrt{x + 1} + 1) } \)
The denominator becomes \( (\sqrt{x+1})^2 - 1^2 = (x+1) - 1 = x \):
\( = \lim_{x \to 0} \frac { (3^x - 1) ( \sqrt{x + 1} + 1 ) }{ x } \)
Now, rearrange the terms to separate the known limit form:
\( = \lim_{x \to 0} \frac { (3^x - 1) }{ x } \times ( \sqrt{x + 1} + 1 ) \)
Apply the limit to each part:
\( = \left( \lim_{x \to 0} \frac { 3^x - 1 }{ x } \right) \times \left( \lim_{x \to 0} ( \sqrt{x + 1} + 1 ) \right) \)
Using the formula for the first limit and substituting \( x = 0 \) into the second part:
\( = ( \log 3 ) \times ( \sqrt{0 + 1} + 1 ) \)
\( = ( \log 3 ) \times ( \sqrt{1} + 1 ) \)
\( = ( \log 3 ) \times ( 1 + 1 ) \)
\( = ( \log 3 ) \times 2 \)
Using the logarithm property \( k \log A = \log A^k \):
\( = \log 3^2 \)
\( = \log 9 \)
The final answer is \( \log 9 \), which shows how rationalizing the denominator can simplify the limit problem.
In simple words: First, we got rid of the square root at the bottom by multiplying it with a special helper term. This made the bottom part just \( x \). Then, we split the problem into two smaller parts that we knew how to solve. We used a special rule for one part and just put in \( x = 0 \) for the other.

🎯 Exam Tip: When limits involve square roots in the denominator (or numerator), always consider rationalizing by multiplying by the conjugate. This often helps in simplifying the expression to apply standard limit formulas.

 

Question 17. Evaluate the following limits:
\( \lim_{x \to 0} \frac { 1 - \cos^2 x }{ x \sin 2x } \)
Answer: We know the trigonometric identities: \( \sin^2 x + \cos^2 x = 1 \implies 1 - \cos^2 x = \sin^2 x \) and \( \sin 2x = 2 \sin x \cos x \). We also know \( \lim_{x \to 0} \frac { \sin x }{ x } = 1 \).
Let's substitute these identities into the limit:
\( \lim_{x \to 0} \frac { 1 - \cos^2 x }{ x \sin 2x } \)
Replace \( 1 - \cos^2 x \) with \( \sin^2 x \) and \( \sin 2x \) with \( 2 \sin x \cos x \):
\( = \lim_{x \to 0} \frac { \sin^2 x }{ x (2 \sin x \cos x) } \)
Cancel one \( \sin x \) term from the numerator and denominator:
\( = \lim_{x \to 0} \frac { \sin x }{ 2x \cos x } \)
Now, separate the terms to match the known limit form \( \frac { \sin x }{ x } \):
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times \frac { 1 }{ 2 \cos x } \right) \)
Apply the limit to each part:
\( = \left( \lim_{x \to 0} \frac { \sin x }{ x } \right) \times \left( \lim_{x \to 0} \frac { 1 }{ 2 \cos x } \right) \)
Substitute the known limit value \( \lim_{x \to 0} \frac { \sin x }{ x } = 1 \) and \( x = 0 \) into the second part (since \( \cos 0 = 1 \)):
\( = 1 \times \frac { 1 }{ 2 \cos 0 } \)
\( = 1 \times \frac { 1 }{ 2 \times 1 } \)
\( = \frac { 1 }{ 2 } \)
Thus, the limit evaluates to \( \frac { 1 }{ 2 } \). This problem highlights the importance of using trigonometric identities for simplification.
In simple words: We used two helper math rules to change the top and bottom parts of the fraction. After simplifying, we arranged the terms so that we could use a known limit rule where \( \sin x \) divided by \( x \) becomes 1. Finally, we put \( x = 0 \) into the rest of the fraction to get the answer.

🎯 Exam Tip: Always look for opportunities to use fundamental trigonometric identities (like \( \sin^2 x + \cos^2 x = 1 \) or double angle formulas) and the standard limit \( \lim_{x \to 0} \frac { \sin x }{ x } = 1 \) to simplify complex trigonometric limit problems.

 

Question 18. Evaluate the following limits:
\( \lim_{x \to \infty} x \left[ 3^{\frac { 1 }{ x }} + 1 - \cos \left( \frac { 1 }{ x } \right) - e^{\frac { 1 }{ x }} \right] \)
Answer: This limit involves \( x \to \infty \), which often suggests a substitution. Let \( y = \frac { 1 }{ x } \).
As \( x \to \infty \), \( y \to 0 \). Also, \( x = \frac { 1 }{ y } \).
Substitute \( y \) into the expression:
\( \lim_{y \to 0} \frac { 1 }{ y } \left[ 3^y + 1 - \cos y - e^y \right] \)
\( = \lim_{y \to 0} \frac { 3^y - e^y + 1 - \cos y }{ y } \)
We can rearrange the terms and split them to use known limit formulas like \( \lim_{y \to 0} \frac { a^y - 1 }{ y } = \log a \), \( \lim_{y \to 0} \frac { e^y - 1 }{ y } = 1 \), and \( \lim_{y \to 0} \frac { 1 - \cos y }{ y } = 0 \).
To do this, we rewrite the numerator by adding and subtracting \( 1 \):
\( = \lim_{y \to 0} \frac { (3^y - 1) - (e^y - 1) + (1 - \cos y) }{ y } \)
Now, separate the fraction into individual terms:
\( = \lim_{y \to 0} \left[ \frac { 3^y - 1 }{ y } - \frac { e^y - 1 }{ y } + \frac { 1 - \cos y }{ y } \right] \)
Apply the limit to each term:
\( = \lim_{y \to 0} \frac { 3^y - 1 }{ y } - \lim_{y \to 0} \frac { e^y - 1 }{ y } + \lim_{y \to 0} \frac { 1 - \cos y }{ y } \)
Using the standard limit formulas:
\( = \log 3 - 1 + 0 \)
\( = \log 3 - 1 \)
The limit evaluates to \( \log 3 - 1 \). This problem shows how a clever substitution and algebraic manipulation can simplify a complex limit into a solvable form.
In simple words: We first changed the problem by replacing \( 1/x \) with \( y \). This made the limit easier to handle because \( y \) now went to zero. Then, we rearranged the top part of the fraction and split it into three smaller parts that matched rules we already knew. We then used those rules to find the answer.

🎯 Exam Tip: For limits as \( x \to \infty \) involving terms like \( 1/x \), a common and effective strategy is to substitute \( y = 1/x \) (so \( y \to 0 \)). Then, manipulate the expression to use standard limits as \( y \to 0 \), such as those for \( a^y \), \( e^y \), or \( \cos y \).

 

Question 19. Evaluate the following limits:
\( \lim_{x \to \infty} \{ x [\log (x + a) - \log(x)] \} \)
Answer: This limit involves logarithms and \( x \to \infty \). We'll use logarithm properties and a substitution to solve it.
The given limit is:
\( \lim_{x \to \infty} \{ x [\log (x + a) - \log(x)] \} \)
First, apply the logarithm property \( \log A - \log B = \log \left( \frac { A }{ B } \right) \):
\( = \lim_{x \to \infty} \left\{ x \log \left( \frac { x + a }{ x } \right) \right\} \)
Simplify the fraction inside the logarithm:
\( = \lim_{x \to \infty} \left\{ x \log \left( 1 + \frac { a }{ x } \right) \right\} \)
Now, let \( y = \frac { a }{ x } \). As \( x \to \infty \), \( y \to 0 \). Also, \( x = \frac { a }{ y } \).
Substitute \( y \) into the expression:
\( = \lim_{y \to 0} \frac { a }{ y } \log (1 + y) \)
Rearrange the terms to match the standard limit \( \lim_{y \to 0} \frac { \log(1 + y) }{ y } = 1 \):
\( = a \lim_{y \to 0} \frac { \log (1 + y) }{ y } \)
Apply the known limit value:
\( = a \times 1 \)
\( = a \)
The limit evaluates to \( a \). This problem beautifully demonstrates the power of logarithmic properties and appropriate substitutions to simplify complex limits.
In simple words: We used a rule for logarithms to combine the two log terms into one. Then, we made a simple change where \( a/x \) became \( y \). This made the problem look like a known limit rule for logarithms that simplifies to 1. Finally, we multiplied by \( a \) to get the answer.

🎯 Exam Tip: When dealing with limits involving \( \log(x+a) - \log x \) as \( x \to \infty \), remember to first combine them using \( \log(A) - \log(B) = \log(A/B) \), which usually leads to \( \log(1 + \frac{a}{x}) \). Then, a substitution like \( y = \frac{a}{x} \) will typically transform it into the standard form \( \lim_{y \to 0} \frac{\log(1+y)}{y} = 1 \).

 

Question 20. Evaluate the following limits:
\( \lim_{x \to \pi} \frac { \sin 3x }{ \sin 2x } \)
Answer: If we directly substitute \( x = \pi \), we get \( \frac { \sin(3\pi) }{ \sin(2\pi) } = \frac { 0 }{ 0 } \), which is an indeterminate form. We need to use trigonometric identities or L'Hopital's rule. The solution uses trigonometric identities.
We know \( \sin 3x = 3 \sin x - 4 \sin^3 x \) and \( \sin 2x = 2 \sin x \cos x \).
Substitute these identities into the limit expression:
\( \lim_{x \to \pi} \frac { 3 \sin x - 4 \sin^3 x }{ 2 \sin x \cos x } \)
Factor out \( \sin x \) from the numerator:
\( = \lim_{x \to \pi} \frac { \sin x (3 - 4 \sin^2 x) }{ 2 \sin x \cos x } \)
Since \( x \to \pi \), \( x \neq \pi \), so \( \sin x \neq 0 \). We can cancel \( \sin x \) from numerator and denominator:
\( = \lim_{x \to \pi} \frac { 3 - 4 \sin^2 x }{ 2 \cos x } \)
Now, substitute \( x = \pi \) into the simplified expression:
\( = \frac { 3 - 4 \sin^2 (\pi) }{ 2 \cos (\pi) } \)
We know \( \sin \pi = 0 \) and \( \cos \pi = -1 \).
\( = \frac { 3 - 4 (0)^2 }{ 2 (-1) } \)
\( = \frac { 3 - 0 }{ -2 } \)
\( = \frac { 3 }{ -2 } \)
Thus, the limit evaluates to \( - \frac { 3 }{ 2 } \). This problem shows how identity application is crucial when direct substitution results in an indeterminate form.
In simple words: When we first tried to put \( x = \pi \) into the problem, we got \( 0/0 \), which means we couldn't find the answer directly. So, we used special math rules for \( \sin 3x \) and \( \sin 2x \) to rewrite the fraction. After canceling out a common term, we put \( x = \pi \) back in and found the answer.

🎯 Exam Tip: When direct substitution into a trigonometric limit yields an indeterminate form like \( 0/0 \), always try to simplify the expression using trigonometric identities before applying the limit. Look for identities that help factor and cancel common terms.

 

Question 21. Evaluate the following limits:
\( \lim_{x \to \frac { \pi }{ 2 }} (1 + \sin x)^{2 \operatorname{cosec} x} \)
Answer: This limit is of the form \( 1^\infty \) as \( x \to \frac { \pi }{ 2 } \) (since \( 1 + \sin (\frac{\pi}{2}) = 1 + 1 = 2 \), and \( 2 \operatorname{cosec} (\frac{\pi}{2}) = 2/1 = 2 \)). Wait, \( \operatorname{cosec} x = 1/\sin x \). As \( x \to \frac{\pi}{2} \), \( \sin x \to 1 \), so \( (1+\sin x) \to (1+1) = 2 \). And \( 2 \operatorname{cosec} x \to 2/1 = 2 \). This means the limit is actually \( 2^2 = 4 \). Let's re-examine the given problem and solution carefully.
Ah, I made a mistake in the quick check. As \( x \to \frac{\pi}{2} \), \( \sin x \to 1 \). So \( (1 + \sin x) \to (1+1) = 2 \). And \( 2 \operatorname{cosec} x = \frac{2}{\sin x} \to \frac{2}{1} = 2 \). So it seems the direct substitution gives \( 2^2 = 4 \). The solution given in the source follows a path for a \( 1^\infty \) form. Let's trace it and see if it implies the original expression was meant to be different or if there's a subtle point.
The solution performs a substitution: "Put \( y = \sin x \). When \( x \to \frac{\pi}{2} \), then \( y \to \sin (\frac{\pi}{2}) = 1 \)."
The limit becomes \( \lim_{y \to 1} (1 + y)^{2 \operatorname{cosec} (\sin^{-1} y)} \). This doesn't look like \( (1+y)^{2/y} \) for \( y \to 0 \). Let's re-read the original image.
The image for Q21 clearly shows \( (1 + \sin x)^{2 \operatorname{cosec} x} \) as \( x \to \frac{\pi}{2} \). The solution for Q21 has: "Put \( y = \sin x \). When \( x \to \frac{\pi}{2} \), then \( y = \sin (\frac{\pi}{2}) = 1 \)." This is correct.
Then it shows: \( \lim_{x \to \frac{\pi}{2}} (1 + \sin x)^{2 \operatorname{cosec} x} = \lim_{y \to 1} (1 + y)^{2/y} \). This substitution is incorrect. If \( y = \sin x \), then \( \operatorname{cosec} x = 1/y \). So the exponent should be \( 2/y \). The base is \( (1+y) \). So it should be \( \lim_{y \to 1} (1+y)^{2/y} \). This is what is written in the source. Now, direct substitution into this *new* expression: \( (1+1)^{2/1} = 2^2 = 4 \). This matches my initial direct evaluation. The " \( 1^\infty \) form" logic in my thought process was mistaken for this specific expression, which can be directly evaluated. The solution seems correct, but it unnecessarily uses a substitution. Let's just present it cleanly.

Given the limit:
\( \lim_{x \to \frac { \pi }{ 2 }} (1 + \sin x)^{2 \operatorname{cosec} x} \)
Let's substitute \( x = \frac { \pi }{ 2 } \) directly into the expression:
\( = (1 + \sin \left( \frac { \pi }{ 2 } \right))^{2 \operatorname{cosec} \left( \frac { \pi }{ 2 } \right)} \)
We know that \( \sin \left( \frac { \pi }{ 2 } \right) = 1 \) and \( \operatorname{cosec} \left( \frac { \pi }{ 2 } \right) = \frac { 1 }{ \sin \left( \frac { \pi }{ 2 } \right) } = \frac { 1 }{ 1 } = 1 \).
Substitute these values:
\( = (1 + 1)^{2 \times 1} \)
\( = (2)^2 \)
\( = 4 \)
Alternatively, following the method shown in the source, which is also valid:
Let \( y = \sin x \).
As \( x \to \frac { \pi }{ 2 } \), then \( y \to \sin \left( \frac { \pi }{ 2 } \right) = 1 \).
Also, \( \operatorname{cosec} x = \frac { 1 }{ \sin x } = \frac { 1 }{ y } \).
Substitute these into the limit expression:
\( \lim_{x \to \frac { \pi }{ 2 }} (1 + \sin x)^{2 \operatorname{cosec} x} = \lim_{y \to 1} (1 + y)^{\frac { 2 }{ y }} \)
Now, substitute \( y = 1 \) into this expression:
\( = (1 + 1)^{\frac { 2 }{ 1 }} \)
\( = (2)^2 \)
\( = 4 \)
Both methods yield the same result, \( 4 \). This problem emphasizes understanding the behavior of trigonometric functions at specific points. The substitution method is shown in the source and leads to the correct result, even if direct evaluation was simpler here.
In simple words: We found what \( \sin x \) and \( \operatorname{cosec} x \) become when \( x \) is close to \( \pi/2 \). We then put these numbers into the expression. This led to \( (1+1) \) raised to the power of \( 2 \), which gives \( 4 \).

🎯 Exam Tip: Before trying complex limit techniques, always attempt direct substitution first. If it yields a definite value (not an indeterminate form like \( 0/0 \) or \( \infty/\infty \)), that is your answer. If it's indeterminate, then proceed with other methods like L'Hopital's rule, factorization, or substitutions.

 

Question 22. Evaluate the following limits:
\( \lim_{x \to 0} \frac { \sqrt{2} - \sqrt{1 + \cos x} }{ \sin^2 x } \)
Answer: If we directly substitute \( x = 0 \), we get \( \frac { \sqrt{2} - \sqrt{1 + \cos 0} }{ \sin^2 0 } = \frac { \sqrt{2} - \sqrt{1 + 1} }{ 0 } = \frac { \sqrt{2} - \sqrt{2} }{ 0 } = \frac { 0 }{ 0 } \), an indeterminate form.
We need to rationalize the numerator by multiplying by its conjugate, \( \sqrt{2} + \sqrt{1 + \cos x} \).
\( \lim_{x \to 0} \frac { \sqrt{2} - \sqrt{1 + \cos x} }{ \sin^2 x } \)
Multiply numerator and denominator by \( \sqrt{2} + \sqrt{1 + \cos x} \):
\( = \lim_{x \to 0} \frac { (\sqrt{2} - \sqrt{1 + \cos x}) (\sqrt{2} + \sqrt{1 + \cos x}) }{ \sin^2 x (\sqrt{2} + \sqrt{1 + \cos x}) } \)
The numerator becomes \( (\sqrt{2})^2 - (\sqrt{1 + \cos x})^2 = 2 - (1 + \cos x) = 2 - 1 - \cos x = 1 - \cos x \):
\( = \lim_{x \to 0} \frac { 1 - \cos x }{ \sin^2 x (\sqrt{2} + \sqrt{1 + \cos x}) } \)
Now, use the identity \( \sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x) \):
\( = \lim_{x \to 0} \frac { 1 - \cos x }{ (1 - \cos x)(1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x}) } \)
Cancel \( (1 - \cos x) \) from numerator and denominator (since \( x \to 0 \), \( x \neq 0 \), so \( 1 - \cos x \neq 0 \)):
\( = \lim_{x \to 0} \frac { 1 }{ (1 + \cos x) (\sqrt{2} + \sqrt{1 + \cos x}) } \)
Now, substitute \( x = 0 \):
\( = \frac { 1 }{ (1 + \cos 0) (\sqrt{2} + \sqrt{1 + \cos 0}) } \)
Since \( \cos 0 = 1 \):
\( = \frac { 1 }{ (1 + 1) (\sqrt{2} + \sqrt{1 + 1}) } \)
\( = \frac { 1 }{ (2) (\sqrt{2} + \sqrt{2}) } \)
\( = \frac { 1 }{ 2 (2\sqrt{2}) } \)
\( = \frac { 1 }{ 4\sqrt{2} } \)
The final answer is \( \frac { 1 }{ 4\sqrt{2} } \). This problem shows the combined use of rationalization and trigonometric identities to evaluate limits.
In simple words: First, we made the top part of the fraction simpler by multiplying it with its opposite. This removed the square roots from the top. Next, we used a known math rule to change \( \sin^2 x \) at the bottom. We then canceled common parts and put \( x = 0 \) into the remaining expression to get the answer.

🎯 Exam Tip: For limits involving square roots and trigonometric functions, always try to rationalize the expression first. After rationalizing, look for trigonometric identities (especially \( \sin^2 x = 1 - \cos^2 x \)) to further simplify the terms and cancel out factors that lead to the indeterminate form.

 

Question 23. Evaluate the following limits:
\( \lim_{x \to 0} \frac { \sqrt{1 + \sin x} - \sqrt{1 - \sin x} }{ \tan x } \)
Answer: If we directly substitute \( x = 0 \), we get \( \frac { \sqrt{1 + \sin 0} - \sqrt{1 - \sin 0} }{ \tan 0 } = \frac { \sqrt{1} - \sqrt{1} }{ 0 } = \frac { 0 }{ 0 } \), an indeterminate form.
We need to rationalize the numerator by multiplying by its conjugate, \( \sqrt{1 + \sin x} + \sqrt{1 - \sin x} \).
\( \lim_{x \to 0} \frac { \sqrt{1 + \sin x} - \sqrt{1 - \sin x} }{ \tan x } \)
Multiply numerator and denominator by \( \sqrt{1 + \sin x} + \sqrt{1 - \sin x} \):
\( = \lim_{x \to 0} \frac { (\sqrt{1 + \sin x} - \sqrt{1 - \sin x}) (\sqrt{1 + \sin x} + \sqrt{1 - \sin x}) }{ \tan x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x}) } \)
The numerator becomes \( (1 + \sin x) - (1 - \sin x) = 1 + \sin x - 1 + \sin x = 2 \sin x \):
\( = \lim_{x \to 0} \frac { 2 \sin x }{ \tan x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x}) } \)
Now, replace \( \tan x \) with \( \frac { \sin x }{ \cos x } \):
\( = \lim_{x \to 0} \frac { 2 \sin x }{ \frac { \sin x }{ \cos x } (\sqrt{1 + \sin x} + \sqrt{1 - \sin x}) } \)
Cancel \( \sin x \) from numerator and denominator (since \( x \to 0 \), \( x \neq 0 \), so \( \sin x \neq 0 \)):
\( = \lim_{x \to 0} \frac { 2 }{ \frac { 1 }{ \cos x } (\sqrt{1 + \sin x} + \sqrt{1 - \sin x}) } \)
Simplify the fraction (multiplying the numerator by \( \cos x \)):
\( = \lim_{x \to 0} \frac { 2 \cos x }{ \sqrt{1 + \sin x} + \sqrt{1 - \sin x} } \)
Now, substitute \( x = 0 \):
\( = \frac { 2 \cos 0 }{ \sqrt{1 + \sin 0} + \sqrt{1 - \sin 0} } \)
Since \( \cos 0 = 1 \) and \( \sin 0 = 0 \):
\( = \frac { 2 \times 1 }{ \sqrt{1 + 0} + \sqrt{1 - 0} } \)
\( = \frac { 2 }{ \sqrt{1} + \sqrt{1} } \)
\( = \frac { 2 }{ 1 + 1 } \)
\( = \frac { 2 }{ 2 } \)
\( = 1 \)
The limit evaluates to \( 1 \). This problem shows how rationalization, combined with basic trigonometric substitutions, can simplify complex limit expressions.
In simple words: First, we made the top part simpler by multiplying it by its special opposite, which removed the square roots. Then, we replaced \( \tan x \) with \( \sin x \) over \( \cos x \). After canceling out common terms, we put \( x = 0 \) into the expression to find the final answer.

🎯 Exam Tip: When limits involve square roots with trigonometric functions, rationalizing the numerator is usually the first step. After that, use fundamental identities like \( \tan x = \sin x / \cos x \) to simplify and cancel terms, leading to a direct substitution.

 

Question 24. Evaluate the following limits:
\( \lim_{x \to \infty} \left( \frac { x^2 - 2x + 1 }{ x^2 - 4x + 2 } \right)^x \)
Answer: This limit is of the form \( 1^\infty \). When \( x \to \infty \), the base \( \frac { x^2 - 2x + 1 }{ x^2 - 4x + 2 } \) approaches \( 1 \) (since the highest powers of \( x \) are the same in numerator and denominator), and the exponent \( x \) approaches \( \infty \).
For limits of the form \( \lim_{x \to a} (f(x))^{g(x)} \) where \( f(x) \to 1 \) and \( g(x) \to \infty \), the limit is \( e^{\lim_{x \to a} (f(x) - 1) g(x)} \).
First, let's find \( f(x) - 1 \):
\( f(x) - 1 = \frac { x^2 - 2x + 1 }{ x^2 - 4x + 2 } - 1 \)
\( = \frac { (x^2 - 2x + 1) - (x^2 - 4x + 2) }{ x^2 - 4x + 2 } \)
\( = \frac { x^2 - 2x + 1 - x^2 + 4x - 2 }{ x^2 - 4x + 2 } \)
\( = \frac { 2x - 1 }{ x^2 - 4x + 2 } \)
Now, we need to calculate \( \lim_{x \to \infty} (f(x) - 1) g(x) \):
\( \lim_{x \to \infty} \left( \frac { 2x - 1 }{ x^2 - 4x + 2 } \right) (x) \)
\( = \lim_{x \to \infty} \frac { x(2x - 1) }{ x^2 - 4x + 2 } \)
\( = \lim_{x \to \infty} \frac { 2x^2 - x }{ x^2 - 4x + 2 } \)
To evaluate this limit as \( x \to \infty \), divide the numerator and denominator by the highest power of \( x \), which is \( x^2 \):
\( = \lim_{x \to \infty} \frac { \frac { 2x^2 }{ x^2 } - \frac { x }{ x^2 } }{ \frac { x^2 }{ x^2 } - \frac { 4x }{ x^2 } + \frac { 2 }{ x^2 } } \)
\( = \lim_{x \to \infty} \frac { 2 - \frac { 1 }{ x } }{ 1 - \frac { 4 }{ x } + \frac { 2 }{ x^2 } } \)
As \( x \to \infty \), terms like \( \frac { 1 }{ x } \) and \( \frac { 2 }{ x^2 } \) go to \( 0 \):
\( = \frac { 2 - 0 }{ 1 - 0 + 0 } \)
\( = 2 \)
Therefore, the original limit is \( e^2 \). This method is a standard technique for evaluating indeterminate forms of type \( 1^\infty \).
In simple words: This problem was of a special type where the base goes to 1 and the power goes to infinity. We used a rule that says such limits are \( e \) raised to a new limit. We calculated that new limit by simplifying the fraction and then dividing everything by the highest power of \( x \). The result was \( e \) to the power of 2.

🎯 Exam Tip: For limits of the form \( 1^\infty \), always remember the formula \( \lim_{x \to a} (f(x))^{g(x)} = e^{\lim_{x \to a} (f(x) - 1) g(x)} \). The crucial first step is to correctly calculate \( f(x) - 1 \) and then evaluate the exponent's limit by dividing by the highest power of \( x \) if \( x \to \infty \).

 

Question 25. Evaluate the following limits:
\( \lim_{x \to 0} \frac { e^x - e^{-x} }{ \sin x } \)
Answer: If we directly substitute \( x = 0 \), we get \( \frac { e^0 - e^0 }{ \sin 0 } = \frac { 1 - 1 }{ 0 } = \frac { 0 }{ 0 } \), an indeterminate form.
We know the standard limits: \( \lim_{x \to 0} \frac { e^x - 1 }{ x } = 1 \) and \( \lim_{x \to 0} \frac { \sin x }{ x } = 1 \).
We can rewrite the numerator to use these standard forms.
\( \lim_{x \to 0} \frac { e^x - e^{-x} }{ \sin x } \)
Multiply and divide by \( x \) to separate the terms into known limit forms:
\( = \lim_{x \to 0} \frac { (e^x - e^{-x}) / x }{ (\sin x) / x } \)
Now, focus on the numerator: \( \frac { e^x - e^{-x} }{ x } \). We can add and subtract \( 1 \) to create the desired forms:
\( \frac { e^x - 1 - (e^{-x} - 1) }{ x } \)
This can be written as: \( \frac { e^x - 1 }{ x } - \frac { e^{-x} - 1 }{ x } \).
Let's apply this back to the main limit:
\( = \frac { \lim_{x \to 0} \left( \frac { e^x - 1 }{ x } - \frac { e^{-x} - 1 }{ x } \right) }{ \lim_{x \to 0} \frac { \sin x }{ x } } \)
For \( \lim_{x \to 0} \frac { e^{-x} - 1 }{ x } \), let \( u = -x \). As \( x \to 0 \), \( u \to 0 \). Then it becomes \( \lim_{u \to 0} \frac { e^u - 1 }{ -u } = - \lim_{u \to 0} \frac { e^u - 1 }{ u } = -1 \).
So, the numerator limit is \( 1 - (-1) = 1 + 1 = 2 \).
The denominator limit is \( 1 \).
Therefore:
\( = \frac { 1 - (-1) }{ 1 } \)
\( = \frac { 2 }{ 1 } \)
\( = 2 \)
The limit evaluates to \( 2 \). This problem showcases how careful algebraic manipulation, using the "add and subtract 1" trick and standard limits, can solve complex indeterminate forms.
In simple words: When we tried to put \( x=0 \) directly, we got \( 0/0 \). So, we adjusted the top part of the fraction by adding and taking away 1. This allowed us to split it into terms that matched standard limit rules we knew, involving \( e^x \) and \( \sin x \). After applying these rules, the answer was 2.

🎯 Exam Tip: For limits involving exponential terms like \( e^x \) or \( e^{-x} \) with \( x \to 0 \), always try to create the form \( \frac{e^k - 1}{k} \) (which approaches 1 as \( k \to 0 \)) by adding and subtracting 1. Also, don't forget the fundamental trigonometric limit \( \frac{\sin x}{x} \to 1 \) as \( x \to 0 \).

 

Question 26. Evaluate the following limits:
\( \lim_{x \to 0} \frac { e^{ax} - e^{bx} }{ x } \)
Answer: If we directly substitute \( x = 0 \), we get \( \frac { e^0 - e^0 }{ 0 } = \frac { 1 - 1 }{ 0 } = \frac { 0 }{ 0 } \), an indeterminate form.
We know the standard limit: \( \lim_{x \to 0} \frac { e^{kx} - 1 }{ x } = k \).
We can manipulate the numerator to use this standard form.
\( \lim_{x \to 0} \frac { e^{ax} - e^{bx} }{ x } \)
Add and subtract \( 1 \) in the numerator:
\( = \lim_{x \to 0} \frac { e^{ax} - 1 + 1 - e^{bx} }{ x } \)
Group the terms to match the standard form:
\( = \lim_{x \to 0} \left( \frac { e^{ax} - 1 }{ x } - \frac { e^{bx} - 1 }{ x } \right) \)
Apply the limit to each term:
\( = \lim_{x \to 0} \frac { e^{ax} - 1 }{ x } - \lim_{x \to 0} \frac { e^{bx} - 1 }{ x } \)
Using the standard limit \( \lim_{x \to 0} \frac { e^{kx} - 1 }{ x } = k \), with \( k=a \) for the first term and \( k=b \) for the second term:
\( = a - b \)
The limit evaluates to \( a - b \). This is a common and important limit formula derived directly from the basic exponential limit.
In simple words: We saw that putting \( x=0 \) directly gave us \( 0/0 \). So, we added and subtracted 1 to the top part of the fraction. This helped us split the problem into two parts that looked like a known limit rule for \( e \) raised to a power. We applied that rule to each part, and the answer was simply \( a \) minus \( b \).

🎯 Exam Tip: For limits of the form \( \lim_{x \to 0} \frac{e^{ax} - e^{bx}}{x} \), immediately recognize it as \( a - b \). This is a direct application of splitting the numerator as \( (e^{ax}-1) - (e^{bx}-1) \) and using the fundamental limit \( \lim_{u \to 0} \frac{e^u-1}{u} = 1 \).

 

Question 27. Evaluate the following limits:
\( \lim_{x \to 0} \frac { \sin x (1 - \cos x) }{ x^3 } \)
Answer: If we directly substitute \( x = 0 \), we get \( \frac { \sin 0 (1 - \cos 0) }{ 0^3 } = \frac { 0 (1 - 1) }{ 0 } = \frac { 0 }{ 0 } \), an indeterminate form.
We know the trigonometric identity: \( 1 - \cos x = 2 \sin^2 \left( \frac { x }{ 2 } \right) \). We also know \( \lim_{u \to 0} \frac { \sin u }{ u } = 1 \).
Substitute the identity into the limit expression:
\( \lim_{x \to 0} \frac { \sin x \left( 2 \sin^2 \left( \frac { x }{ 2 } \right) \right) }{ x^3 } \)
Rearrange the terms to group them into the standard \( \frac { \sin u }{ u } \) form:
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times \frac { 2 \sin^2 \left( \frac { x }{ 2 } \right) }{ x^2 } \right) \)
Further separate the \( \frac { \sin^2 \left( \frac { x }{ 2 } \right) }{ x^2 } \) term to match the standard form:
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times 2 \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \times \frac { 1 }{ 4 } \right) \)
(Note: we multiplied \( x^2 \) in the denominator by \( \frac{1}{4} \) and \( 2 \) in the numerator by \( 4 \cdot \frac{1}{4} \) to get \( (\frac{x}{2})^2 \) in the denominator, so the \( 2 \) from \( 1-\cos x \) combines with the new \( 4 \) in the denominator for overall coefficient of \( \frac{1}{2} \)).
Let's do this more systematically:
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times 2 \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ x } \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ x } \right) \)
To make \( \frac { \sin (x/2) }{ x } \) into \( \frac { \sin (x/2) }{ (x/2) } \), we need a \( \frac { 1 }{ 2 } \) in the denominator, so we multiply and divide by \( \frac { 1 }{ 2 } \).
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times 2 \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } \cdot 2 } \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } \cdot 2 } \right) \)
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times 2 \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \times \frac { 1 }{ 2 } \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \times \frac { 1 }{ 2 } \right) \)
Group the terms:
\( = \left( \lim_{x \to 0} \frac { \sin x }{ x } \right) \times 2 \times \left( \lim_{x \to 0} \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \right) \times \left( \lim_{x \to 0} \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \right) \times \frac { 1 }{ 4 } \)
Apply the standard limit \( \lim_{u \to 0} \frac { \sin u }{ u } = 1 \):
\( = 1 \times 2 \times 1 \times 1 \times \frac { 1 }{ 4 } \)
\( = \frac { 2 }{ 4 } \)
\( = \frac { 1 }{ 2 } \)
The limit evaluates to \( \frac { 1 }{ 2 } \). This problem is an excellent example of using trigonometric identities and careful rearrangement to fit standard limit forms.
In simple words: First, we changed \( (1 - \cos x) \) using a known math rule that turned it into \( 2 \sin^2 (x/2) \). Then, we broke the problem into smaller parts, making sure each part looked like \( \sin u / u \). We used the rule that \( \sin u / u \) becomes 1 when \( u \) is very small. After multiplying all the values, we got the answer \( 1/2 \).

🎯 Exam Tip: When \( x^3 \) appears in the denominator alongside \( \sin x \) and \( (1 - \cos x) \) in the numerator, it's a strong hint to use the identity \( 1 - \cos x = 2 \sin^2(x/2) \). Then, carefully adjust the denominators to create three instances of the \( \frac{\sin u}{u} \to 1 \) form.

 

Question 28. Evaluate the following limits:
\( \lim_{x \to 0} \frac { \tan x - \sin x }{ x^3 } \)
Answer: If we directly substitute \( x = 0 \), we get \( \frac { \tan 0 - \sin 0 }{ 0^3 } = \frac { 0 - 0 }{ 0 } = \frac { 0 }{ 0 } \), an indeterminate form.
We know the identities: \( \tan x = \frac { \sin x }{ \cos x } \) and \( 1 - \cos x = 2 \sin^2 \left( \frac { x }{ 2 } \right) \). Also, \( \lim_{u \to 0} \frac { \sin u }{ u } = 1 \).
First, rewrite the numerator \( \tan x - \sin x \):
\( \tan x - \sin x = \frac { \sin x }{ \cos x } - \sin x \)
Factor out \( \sin x \):
\( = \sin x \left( \frac { 1 }{ \cos x } - 1 \right) \)
Combine the terms inside the parenthesis:
\( = \sin x \left( \frac { 1 - \cos x }{ \cos x } \right) \)
Now, substitute this back into the limit expression:
\( \lim_{x \to 0} \frac { \sin x \left( \frac { 1 - \cos x }{ \cos x } \right) }{ x^3 } \)
Rearrange the terms:
\( = \lim_{x \to 0} \frac { \sin x (1 - \cos x) }{ x^3 \cos x } \)
Now, use the identity \( 1 - \cos x = 2 \sin^2 \left( \frac { x }{ 2 } \right) \):
\( = \lim_{x \to 0} \frac { \sin x \left( 2 \sin^2 \left( \frac { x }{ 2 } \right) \right) }{ x^3 \cos x } \)
Group the terms to fit the standard \( \frac { \sin u }{ u } \) form. We need \( x \) for \( \sin x \), and \( (x/2)^2 \) for \( \sin^2(x/2) \). So, the \( x^3 \) in the denominator will be split as \( x \cdot (x/2) \cdot (x/2) \cdot 4 \).
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times \frac { 2 \sin^2 \left( \frac { x }{ 2 } \right) }{ x^2 } \times \frac { 1 }{ \cos x } \right) \)
\( = \lim_{x \to 0} \left( \frac { \sin x }{ x } \times 2 \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \times \frac { 1 }{ 2 } \times \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \times \frac { 1 }{ 2 } \times \frac { 1 }{ \cos x } \right) \)
Apply the limit to each part, using \( \lim_{u \to 0} \frac { \sin u }{ u } = 1 \) and \( \cos 0 = 1 \):
\( = \left( \lim_{x \to 0} \frac { \sin x }{ x } \right) \times 2 \times \left( \lim_{x \to 0} \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \right) \times \frac { 1 }{ 2 } \times \left( \lim_{x \to 0} \frac { \sin \left( \frac { x }{ 2 } \right) }{ \frac { x }{ 2 } } \right) \times \frac { 1 }{ 2 } \times \left( \lim_{x \to 0} \frac { 1 }{ \cos x } \right) \)
\( = 1 \times 2 \times 1 \times \frac { 1 }{ 2 } \times 1 \times \frac { 1 }{ 2 } \times \frac { 1 }{ 1 } \)
\( = 2 \times \frac { 1 }{ 4 } \)
\( = \frac { 1 }{ 2 } \)
The limit evaluates to \( \frac { 1 }{ 2 } \). This problem effectively combines algebraic simplification, trigonometric identities, and the application of standard limit forms.
In simple words: First, we changed \( \tan x \) into \( \sin x / \cos x \). Then, we made the top part simpler by taking \( \sin x \) out and combining the rest. We used a special rule for \( (1 - \cos x) \) and then split the problem into many small parts that each matched the \( \sin u / u \) rule. After putting in the known values, the answer was \( 1/2 \).

🎯 Exam Tip: When \( \tan x - \sin x \) appears in a limit as \( x \to 0 \), always factor out \( \sin x \) first to get \( \sin x (\frac{1}{\cos x} - 1) \), then simplify the parenthesis to \( \sin x \frac{1-\cos x}{\cos x} \). This sets up the problem for using \( 1-\cos x = 2\sin^2(x/2) \) and the fundamental limit \( \frac{\sin u}{u} \to 1 \).

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