OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Chapter Test

Question 1. \( \lim_{x\to \frac{1}{2}} \frac{4x^2-1}{2x-1} \)
Answer: We need to find the limit of the given expression as \( x \) approaches \( \frac{1}{2} \).
\( \text{Sol. } \lim_{x\to \frac{1}{2}} \frac{4x^2-1}{2x-1} \)
We can factor the numerator \( 4x^2 - 1 \) as a difference of squares: \( (2x)^2 - 1^2 = (2x - 1)(2x + 1) \).
This helps us simplify the expression.
\( \implies \lim_{x\to \frac{1}{2}} \frac{(2x - 1)(2x + 1)}{2x - 1} \)
Now, we can cancel out the common term \( (2x - 1) \) from the numerator and denominator, since \( x \neq \frac{1}{2} \) when taking the limit.
\( \implies \lim_{x\to \frac{1}{2}} (2x + 1) \)
Next, substitute \( x = \frac{1}{2} \) into the simplified expression to evaluate the limit.
\( \implies 2\left(\frac{1}{2}\right) + 1 \)
\( \implies 1 + 1 \)
\( \implies 2 \)
In simple words: First, break down the top part of the fraction using a special algebra rule. Then, you can cancel out matching parts from the top and bottom. Finally, put the given number into the simplified sum to get your answer.

🎯 Exam Tip: When dealing with limits, always try to factorize and simplify the expression first, especially if direct substitution results in an indeterminate form (like 0/0).

Question 2. \( \lim_{x\to 0} \frac{\sin 3x}{\sin 2x} \)
Answer: We want to find the limit of the given trigonometric expression as \( x \) approaches 0.
\( \text{Sol. } \lim_{x\to 0} \frac{\sin 3x}{\sin 2x} \)
To solve this, we use the standard limit identity: \( \lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1 \). We need to adjust the expression to fit this form.
\( \implies \lim_{x\to 0} \frac{\sin 3x}{3x} \times \frac{3x}{2x} \times \frac{2x}{\sin 2x} \)
We can rearrange the terms and split the limit of the product into the product of limits.
\( \implies \lim_{x\to 0} \left(\frac{\sin 3x}{3x}\right) \times \left(\frac{3}{2}\right) \times \lim_{x\to 0} \left(\frac{2x}{\sin 2x}\right) \)
As \( x \to 0 \), both \( 3x \to 0 \) and \( 2x \to 0 \). So, we can apply the limit identity.
\( \implies 1 \times \frac{3}{2} \times 1 \)
\( \implies \frac{3}{2} \)
In simple words: To solve this, make the expression look like "sin(something) divided by (that same something)" for both the top and bottom parts. Since that special limit is 1, you can then easily find the answer by multiplying the numbers left over.

🎯 Exam Tip: Remember the fundamental trigonometric limit \( \lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1 \). When solving, multiply and divide by appropriate terms to create this form for both the numerator and the denominator.

Question 3. \( \lim_{x\to 0} \frac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x} \)
Answer: We need to evaluate the limit of the given trigonometric expression as \( x \) approaches 0.
\( \text{Sol. } \lim_{x\to 0} \frac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x} \)
First, we use the sum-to-product and difference-to-product trigonometric formulas. For the numerator, use \( \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \). For the denominator, use \( \sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \).
So, \( \sin 2x + \sin 6x = 2\sin\left(\frac{2x+6x}{2}\right)\cos\left(\frac{2x-6x}{2}\right) = 2\sin(4x)\cos(-2x) \). Remember \( \cos(-\theta) = \cos(\theta) \).
And \( \sin 5x - \sin 3x = 2\cos\left(\frac{5x+3x}{2}\right)\sin\left(\frac{5x-3x}{2}\right) = 2\cos(4x)\sin(x) \).
\( \implies \lim_{x\to 0} \frac{2\sin 4x \cos(2x)}{2\cos 4x \sin x} \)
We can cancel the 2s and rearrange the terms to apply the standard limit identity \( \lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1 \).
\( \implies \lim_{x\to 0} \left(\frac{\sin 4x}{4x} \times 4x\right) \times \frac{\cos 2x}{\cos 4x} \times \frac{1}{\left(\frac{\sin x}{x} \times x\right)} \)
\( \implies \lim_{x\to 0} \frac{\sin 4x}{\sin x} \times \frac{\cos 2x}{\cos 4x} \)
Now, divide the numerator and denominator by \( x \) and rearrange terms to use the limit property.
\( \implies \lim_{x\to 0} \frac{\left(\frac{\sin 4x}{4x}\right) \times 4}{\left(\frac{\sin x}{x}\right) \times 1} \times \lim_{x\to 0} \frac{\cos 2x}{\cos 4x} \)
As \( x \to 0 \), \( \frac{\sin 4x}{4x} \to 1 \) and \( \frac{\sin x}{x} \to 1 \). Also, \( \cos(0) = 1 \).
\( \implies \frac{1 \times 4}{1 \times 1} \times \frac{\cos(0)}{\cos(0)} \)
\( \implies 4 \times \frac{1}{1} \)
\( \implies 4 \)
In simple words: First, use special math rules to change the sums and differences of 'sin' into multiplications. Then, divide both the top and bottom by 'x' and arrange them so you can use the rule that 'sin(x)/x' becomes 1 as 'x' gets very small. Finally, put 0 into any 'cos' parts and calculate the rest.

🎯 Exam Tip: When dealing with sums or differences of trigonometric functions in limits, always consider using sum-to-product or difference-to-product formulas first to simplify the expression. Then, apply the standard limit identities.

Question 4. \( \lim_{x\to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x} \)
Answer: We need to find the limit of the given expression as \( x \) approaches 0.
\( \text{Sol. } \lim_{x\to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x} \)
We will divide both the numerator and the denominator by \( x \) to use the standard limit identities \( \lim_{\theta\to 0} \frac{\tan \theta}{\theta} = 1 \) and \( \lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1 \).
\( \implies \lim_{x\to 0} \frac{\left(\frac{\tan 3x}{x}\right) - \left(\frac{2x}{x}\right)}{\left(\frac{3x}{x}\right) - \left(\frac{\sin^2 x}{x}\right)} \)
Now, we can simplify each term. Note that \( \frac{\sin^2 x}{x} = \frac{\sin x}{x} \times \sin x \).
\( \implies \lim_{x\to 0} \frac{\left(3 \times \frac{\tan 3x}{3x}\right) - 2}{3 - \left(\frac{\sin x}{x} \times \sin x\right)} \)
As \( x \to 0 \), \( \frac{\tan 3x}{3x} \to 1 \) and \( \frac{\sin x}{x} \to 1 \). Also, \( \sin x \to \sin 0 = 0 \).
\( \implies \frac{(3 \times 1) - 2}{3 - (1 \times 0)} \)
\( \implies \frac{3 - 2}{3 - 0} \)
\( \implies \frac{1}{3} \)
In simple words: Divide every part of the top and bottom of the fraction by 'x'. Then, use the special rules that 'tan(something) divided by (something)' and 'sin(something) divided by (something)' both become 1 when that 'something' is very close to zero. Once you do that, the calculation becomes simple.

🎯 Exam Tip: When faced with algebraic-trigonometric expressions in limits as \( x \to 0 \), dividing both numerator and denominator by the highest power of \( x \) (or simply \( x \) if terms like \( \sin x \) or \( \tan x \) are present) is often an effective strategy to apply standard limit formulas.

Question 5. \( \lim_{x\to \infty} \frac{2x^2 + 7x + 5}{4x^2 + 3x - 1} \)
Answer: We need to find the limit of the rational function as \( x \) approaches infinity.
\( \text{Sol. } \lim_{x\to \infty} \frac{2x^2 + 7x + 5}{4x^2 + 3x - 1} \)
To evaluate limits at infinity for rational functions, we divide every term in the numerator and the denominator by the highest power of \( x \) present in the denominator. In this case, the highest power is \( x^2 \).
\( \implies \lim_{x\to \infty} \frac{\frac{2x^2}{x^2} + \frac{7x}{x^2} + \frac{5}{x^2}}{\frac{4x^2}{x^2} + \frac{3x}{x^2} - \frac{1}{x^2}} \)
Simplify the terms.
\( \implies \lim_{x\to \infty} \frac{2 + \frac{7}{x} + \frac{5}{x^2}}{4 + \frac{3}{x} - \frac{1}{x^2}} \)
Now, apply the limit. Remember that as \( x \to \infty \), any term of the form \( \frac{C}{x^n} \) where \( C \) is a constant and \( n > 0 \) approaches 0.
\( \implies \frac{2 + 0 + 0}{4 + 0 - 0} \)
\( \implies \frac{2}{4} \)
\( \implies \frac{1}{2} \)
In simple words: When finding a limit as 'x' gets extremely big, divide every part of the top and bottom of the fraction by the biggest power of 'x' you see. Any term with 'x' in the bottom will then become zero. What's left will be your answer.

🎯 Exam Tip: For limits at infinity of rational functions, always divide every term by the highest power of \( x \) in the denominator. This makes terms with \( x \) in the denominator tend to zero, simplifying the calculation significantly.

Question 6. \( \lim_{x\to 0} \frac{2\sin x - \sin 2x}{x^3} \)
Answer: We need to evaluate the limit of the given trigonometric expression as \( x \) approaches 0.
\( \text{Sol. } \lim_{x\to 0} \frac{2\sin x - \sin 2x}{x^3} \)
First, we use the double angle identity for sine: \( \sin 2x = 2\sin x \cos x \). This will help simplify the numerator.
\( \implies \lim_{x\to 0} \frac{2\sin x - 2\sin x \cos x}{x^3} \)
Now, factor out \( 2\sin x \) from the numerator.
\( \implies \lim_{x\to 0} \frac{2\sin x (1 - \cos x)}{x^3} \)
We know another useful identity: \( 1 - \cos x = 2\sin^2\left(\frac{x}{2}\right) \). Substitute this into the expression.
\( \implies \lim_{x\to 0} \frac{2\sin x \left(2\sin^2\left(\frac{x}{2}\right)\right)}{x^3} \)
Rearrange the terms to group them in a way that allows us to use the standard limit identity \( \lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1 \).
\( \implies \lim_{x\to 0} 2 \times \frac{\sin x}{x} \times \frac{2\sin^2\left(\frac{x}{2}\right)}{x^2} \)
Split \( x^2 \) as \( x/2 \times x/2 \times 4 \) for the \( \sin^2(x/2) \) term.
\( \implies \lim_{x\to 0} 2 \times \frac{\sin x}{x} \times 2 \times \left(\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2 \times \frac{1}{4} \times 4 \)
\( \implies 2 \times \lim_{x\to 0} \frac{\sin x}{x} \times \lim_{x\to 0} \left(\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2 \)
As \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \) and \( \frac{\sin(x/2)}{x/2} \to 1 \).
\( \implies 2 \times 1 \times (1)^2 \)
\( \implies 2 \)
In simple words: First, replace 'sin 2x' with '2 sin x cos x' using a trigonometric rule. Then, take out '2 sin x' as a common factor. Next, change '1 - cos x' using another rule that turns it into 'sin squared'. After that, arrange the terms carefully to use the 'sin(angle)/angle' rule, which becomes 1. Finally, multiply the remaining numbers to get the answer.

🎯 Exam Tip: When \( x \to 0 \) and trigonometric functions are involved, always look for opportunities to use double-angle or half-angle identities to simplify expressions, especially when you see combinations like \( 1 - \cos x \) or \( \sin 2x \).

Question 7. \( \lim_{x\to a} \frac{\sqrt{1+ax} - \sqrt{1-ax}}{x} \)
Answer: We need to find the limit of the given expression as \( x \) approaches \( a \). The problem statement indicates that the limit is not in indeterminate form initially.
\( \text{Sol. } \lim_{x\to a} \frac{\sqrt{1+ax} - \sqrt{1-ax}}{x} \)
If direct substitution of \( x=a \) does not yield an indeterminate form (like 0/0 or \( \infty/\infty \)), we can substitute \( x=a \) directly.
\( \implies \frac{\sqrt{1+a(a)} - \sqrt{1-a(a)}}{a} \)
\( \implies \frac{\sqrt{1+a^2} - \sqrt{1-a^2}}{a} \)
This is the value of the limit. No further simplification is needed unless \( a=0 \), in which case it would become \( \frac{1-1}{0} = \frac{0}{0} \), an indeterminate form that would require rationalization.
If the problem implies \( x \to 0 \) instead of \( x \to a \) (which is a common pattern for this type of question when rationalization is expected), the solution would be different. But strictly following \( x \to a \), and given the note, the direct substitution holds.
If we were to rationalize (assuming \( x \to 0 \) was intended for an indeterminate form):
Multiply by the conjugate of the numerator: \( \frac{\sqrt{1+ax} + \sqrt{1-ax}}{\sqrt{1+ax} + \sqrt{1-ax}} \)
\( \implies \lim_{x\to 0} \frac{(1+ax) - (1-ax)}{x(\sqrt{1+ax} + \sqrt{1-ax})} \)
\( \implies \lim_{x\to 0} \frac{2ax}{x(\sqrt{1+ax} + \sqrt{1-ax})} \)
Cancel \( x \):
\( \implies \lim_{x\to 0} \frac{2a}{\sqrt{1+ax} + \sqrt{1-ax}} \)
Substitute \( x = 0 \):
\( \implies \frac{2a}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2a}{1+1} = \frac{2a}{2} = a \)
However, based on the question statement \( \lim_{x\to a} \) and the note, the initial direct substitution result is the correct approach here.
In simple words: Since this limit does not result in an unclear form (like 0 divided by 0) when you directly put 'a' in place of 'x', you simply replace 'x' with 'a' in the entire expression. The result you get is the final answer.

🎯 Exam Tip: Always first attempt direct substitution to evaluate a limit. If it results in a defined real number, that is your answer. Only resort to techniques like factorization or rationalization if direct substitution leads to an indeterminate form (e.g., \( \frac{0}{0} \), \( \frac{\infty}{\infty} \)).

Question 8. \( \lim_{x\to \pi} \frac{\sin x}{\pi - x} \)
Answer: We need to find the limit of the given expression as \( x \) approaches \( \pi \).
\( \text{Sol. } \lim_{x\to \pi} \frac{\sin x}{\pi - x} \)
If we substitute \( x = \pi \) directly, we get \( \frac{\sin \pi}{\pi - \pi} = \frac{0}{0} \), which is an indeterminate form. We need to use a substitution.
Let \( x = \pi - h \). As \( x \to \pi \), we have \( \pi - h \to \pi \), which means \( h \to 0 \).
When \( x = \pi - h \), then \( \pi - x = \pi - (\pi - h) = h \).
Also, \( \sin x = \sin (\pi - h) \). Using the identity \( \sin(\pi - \theta) = \sin \theta \), we get \( \sin x = \sin h \).
So, the limit becomes:
\( \implies \lim_{h\to 0} \frac{\sin h}{h} \)
This is a standard trigonometric limit, which equals 1.
\( \implies 1 \)
We can also verify this using L.H.L and R.H.L.
L.H.L (Left Hand Limit): Put \( x = \pi - h \). As \( x \to \pi^- \), \( h \to 0^+ \).
\( \implies \lim_{h\to 0^+} \frac{\sin(\pi - h)}{\pi - (\pi - h)} = \lim_{h\to 0^+} \frac{\sin h}{h} = 1 \)
R.H.L (Right Hand Limit): Put \( x = \pi + h \). As \( x \to \pi^+ \), \( h \to 0^+ \).
\( \implies \lim_{h\to 0^+} \frac{\sin(\pi + h)}{\pi - (\pi + h)} = \lim_{h\to 0^+} \frac{-\sin h}{-h} = \lim_{h\to 0^+} \frac{\sin h}{h} = 1 \)
Since L.H.L = R.H.L = 1, the limit exists and is 1.
In simple words: If you try to put 'pi' into the 'x' directly, you get zero divided by zero, which is not allowed. To fix this, replace 'x' with 'pi minus h'. As 'x' gets close to 'pi', 'h' will get close to zero. Use a sine rule to change 'sin(pi - h)' to 'sin h', and then you'll see a very common limit form that always equals 1.

🎯 Exam Tip: For limits involving \( \sin x \) and terms like \( (\pi - x) \) as \( x \to \pi \), a good strategy is to substitute \( x = \pi - h \) (or \( x = \pi + h \)) and use trigonometric identities to convert the limit into the standard form \( \lim_{h\to 0} \frac{\sin h}{h} = 1 \).

Question 9. \( \lim_{x\to 0} \frac{3^{2x} - 1}{2^{3x} - 1} \)
Answer: We need to evaluate the limit of the given exponential expression as \( x \) approaches 0.
\( \text{Sol. } \lim_{x\to 0} \frac{3^{2x} - 1}{2^{3x} - 1} \)
If we substitute \( x = 0 \) directly, we get \( \frac{3^0 - 1}{2^0 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \), which is an indeterminate form. We need to use a standard limit identity for exponential functions: \( \lim_{\theta\to 0} \frac{a^{\theta} - 1}{\theta} = \log a \).
To apply this, we divide both the numerator and the denominator by \( x \).
\( \implies \lim_{x\to 0} \frac{\frac{3^{2x} - 1}{x}}{\frac{2^{3x} - 1}{x}} \)
Now, we adjust the terms in the numerator and denominator to match the form \( \frac{a^{\theta} - 1}{\theta} \).
For the numerator: \( \frac{3^{2x} - 1}{x} = \frac{3^{2x} - 1}{2x} \times 2 \).
For the denominator: \( \frac{2^{3x} - 1}{x} = \frac{2^{3x} - 1}{3x} \times 3 \).
\( \implies \lim_{x\to 0} \frac{\left(\frac{3^{2x} - 1}{2x}\right) \times 2}{\left(\frac{2^{3x} - 1}{3x}\right) \times 3} \)
As \( x \to 0 \), both \( 2x \to 0 \) and \( 3x \to 0 \). So, we can apply the identity.
\( \implies \frac{(\log 3) \times 2}{(\log 2) \times 3} \)
\( \implies \frac{2 \log 3}{3 \log 2} \)
Using the logarithm property \( k \log a = \log a^k \):
\( \implies \frac{\log 3^2}{\log 2^3} = \frac{\log 9}{\log 8} \)
In simple words: When you have numbers raised to 'x' powers, subtract 1, and 'x' goes to zero, you need to use a special limit rule involving 'log'. Divide both the top and bottom of the fraction by 'x', and then multiply by numbers so that the 'x' in the denominator matches the power. This turns each part into a 'log' term, and you can then simplify the result.

🎯 Exam Tip: Remember the exponential limit identity \( \lim_{x\to 0} \frac{a^x - 1}{x} = \log a \). When you see expressions with \( a^{kx} - 1 \), ensure you multiply and divide by \( k \) in the denominator to match the form \( \frac{a^{kx} - 1}{kx} \).

Question 10. \( \lim_{x\to 0} \frac{3^{2x} - 2^{3x}}{x} \)
Answer: We need to evaluate the limit of the given exponential expression as \( x \) approaches 0.
\( \text{Sol. } \lim_{x\to 0} \frac{3^{2x} - 2^{3x}}{x} \)
If we substitute \( x = 0 \) directly, we get \( \frac{3^0 - 2^0}{0} = \frac{1 - 1}{0} = \frac{0}{0} \), which is an indeterminate form. We need to use the standard limit identity for exponential functions: \( \lim_{\theta\to 0} \frac{a^{\theta} - 1}{\theta} = \log a \).
To apply this identity, we need to introduce ' - 1 ' and ' + 1 ' in the numerator to create the desired form.
\( \implies \lim_{x\to 0} \frac{3^{2x} - 1 + 1 - 2^{3x}}{x} \)
Rearrange the terms to group them.
\( \implies \lim_{x\to 0} \frac{(3^{2x} - 1) - (2^{3x} - 1)}{x} \)
Now, we can split this into two separate limits, applying the limit to each term.
\( \implies \lim_{x\to 0} \frac{3^{2x} - 1}{x} - \lim_{x\to 0} \frac{2^{3x} - 1}{x} \)
For each limit, we adjust the denominator to match the exponent. For the first term, multiply and divide by 2. For the second term, multiply and divide by 3.
\( \implies \lim_{x\to 0} \frac{3^{2x} - 1}{2x} \times 2 - \lim_{x\to 0} \frac{2^{3x} - 1}{3x} \times 3 \)
As \( x \to 0 \), both \( 2x \to 0 \) and \( 3x \to 0 \). So, we can apply the identity \( \lim_{\theta\to 0} \frac{a^{\theta} - 1}{\theta} = \log a \).
\( \implies (\log 3) \times 2 - (\log 2) \times 3 \)
\( \implies 2 \log 3 - 3 \log 2 \)
Using the logarithm property \( k \log a = \log a^k \):
\( \implies \log 3^2 - \log 2^3 \)
\( \implies \log 9 - \log 8 \)
Using the logarithm property \( \log a - \log b = \log \left(\frac{a}{b}\right) \):
\( \implies \log \left(\frac{9}{8}\right) \)
In simple words: If you try to put zero into the 'x' directly, you get an unclear answer. To solve this, add and subtract '1' in the top part of the fraction. This lets you separate the problem into two parts, each fitting a special 'log' rule. Adjust the bottom of each part to match the power, then use the rule, and finally combine the 'log' answers.

🎯 Exam Tip: When dealing with limits of expressions like \( a^{kx} - b^{px} \) divided by \( x \) as \( x \to 0 \), always add and subtract 1 in the numerator. This helps you split the expression into standard forms for \( \frac{a^{kx} - 1}{x} \) and \( \frac{b^{px} - 1}{x} \), which can then be evaluated using the \( \log \) identity.