OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Exercise 18 (I)

Question 1.
(i) \( \text{Lim}_{x \to 0} \frac{e^{4x}-1}{x} \)
(ii) \( \text{Lim}_{x \to 0} \frac{3^x-1}{x} \)
(iii) \( \text{Lim}_{x \to 0} \frac{e^x-x-1}{x} \)
Answer:
(i) \( \text{Lim}_{x \to 0} \frac{e^{4x}-1}{x} \)
To solve this, we multiply and divide the expression by 4:
\( = \text{Lim}_{x \to 0} 4 \cdot \frac{e^{4x}-1}{4x} \)
Now, let \( y = 4x \). As \( x \to 0 \), then \( y \to 0 \). The limit becomes:
\( = 4 \cdot \text{Lim}_{y \to 0} \frac{e^y-1}{y} \)
We know the standard limit \( \text{Lim}_{y \to 0} \frac{e^y-1}{y} = \log_e e = 1 \).
So, the limit is \( 4 \cdot 1 = 4 \). This is a standard limit formula \( \lim_{x \to 0} \frac{e^{ax}-1}{x} = a \).
(ii) \( \text{Lim}_{x \to 0} \frac{3^x-1}{x} \)
This is a standard limit of the form \( \text{Lim}_{x \to 0} \frac{a^x-1}{x} = \log_e a \).
Here, \( a = 3 \). So, the limit is \( \log_e 3 \). This limit is important for understanding how exponential functions behave near x=0.
(iii) \( \text{Lim}_{x \to 0} \frac{e^x-x-1}{x} \)
We can split the expression into two parts:
\( = \text{Lim}_{x \to 0} \left( \frac{e^x-1}{x} - \frac{x}{x} \right) \)
\( = \text{Lim}_{x \to 0} \frac{e^x-1}{x} - \text{Lim}_{x \to 0} 1 \)
We know \( \text{Lim}_{x \to 0} \frac{e^x-1}{x} = \log_e e = 1 \).
So, the limit is \( 1 - 1 = 0 \). This shows how a more complex limit can be broken down using known limit properties.
In simple words: For the first part, we used a rule that says when 'e' is raised to a number multiplied by 'x', minus 1, all divided by 'x', the limit is just that number. For the second part, we used a rule for numbers raised to 'x', minus 1, all divided by 'x', which gives the natural logarithm of the number. For the third part, we split the problem into two easier parts and used known rules.

🎯 Exam Tip: Familiarize yourself with the basic exponential and logarithmic limit formulas like \( \lim_{x \to 0} \frac{e^{ax}-1}{x} = a \) and \( \lim_{x \to 0} \frac{a^x-1}{x} = \log_e a \). This helps to quickly solve such problems.

Question 2.
(i) \( \text{Lim}_{x \to 0} \frac{x(e^x-1)}{1-\cos 2x} \)
(ii) \( \text{Lim}_{x \to 0} \frac{x(2^x-1)}{1-\cos x} \)
(iii) \( \text{Lim}_{x \to 0} \frac{e^x-\sin x-1}{x} \)
(iv) \( \text{Lim}_{x \to 0} e^x \)
Answer:
(i) \( \text{Lim}_{x \to 0} \frac{x(e^x-1)}{1-\cos 2x} \)
We use the trigonometric identity \( 1 - \cos 2x = 2\sin^2 x \).
So, the expression becomes \( \text{Lim}_{x \to 0} \frac{x(e^x-1)}{2\sin^2 x} \).
We can rearrange this as:
\( = \text{Lim}_{x \to 0} \frac{e^x-1}{x} \cdot \frac{x^2}{2\sin^2 x} \)
\( = \text{Lim}_{x \to 0} \frac{e^x-1}{x} \cdot \frac{1}{2} \cdot \left(\frac{x}{\sin x}\right)^2 \)
We know the standard limits:
\( \text{Lim}_{x \to 0} \frac{e^x-1}{x} = \log_e e = 1 \)
\( \text{Lim}_{x \to 0} \frac{x}{\sin x} = 1 \)
So, the limit is \( 1 \cdot \frac{1}{2} \cdot (1)^2 = \frac{1}{2} \). This solution skillfully uses trigonometric identities and fundamental limits to simplify a complex expression.
(ii) \( \text{Lim}_{x \to 0} \frac{x(2^x-1)}{1-\cos x} \)
We use the trigonometric identity \( 1 - \cos x = 2\sin^2 \frac{x}{2} \).
So, the expression becomes \( \text{Lim}_{x \to 0} \frac{x(2^x-1)}{2\sin^2 \frac{x}{2}} \).
We can rearrange this as:
\( = \text{Lim}_{x \to 0} \frac{2^x-1}{x} \cdot \frac{x^2}{2\sin^2 \frac{x}{2}} \)
\( = \text{Lim}_{x \to 0} \frac{2^x-1}{x} \cdot \frac{1}{2} \cdot \left(\frac{x}{\sin \frac{x}{2}}\right)^2 \)
For the second part, we can write \( \frac{x}{\sin \frac{x}{2}} = \frac{2 \cdot \frac{x}{2}}{\sin \frac{x}{2}} \).
We know the standard limits:
\( \text{Lim}_{x \to 0} \frac{2^x-1}{x} = \log_e 2 \)
\( \text{Lim}_{y \to 0} \frac{y}{\sin y} = 1 \), so \( \text{Lim}_{x \to 0} \frac{\frac{x}{2}}{\sin \frac{x}{2}} = 1 \).
So, the limit is \( \log_e 2 \cdot \frac{1}{2} \cdot (2 \cdot 1)^2 = \log_e 2 \cdot \frac{1}{2} \cdot 4 = 2 \log_e 2 \).
Using logarithm properties, \( 2 \log_e 2 = \log_e (2^2) = \log_e 4 \). Remember to adjust the argument of the sine function carefully when applying the \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) formula.
(iii) \( \text{Lim}_{x \to 0} \frac{e^x-\sin x-1}{x} \)
We can split the expression into two parts:
\( = \text{Lim}_{x \to 0} \left( \frac{e^x-1}{x} - \frac{\sin x}{x} \right) \)
\( = \text{Lim}_{x \to 0} \frac{e^x-1}{x} - \text{Lim}_{x \to 0} \frac{\sin x}{x} \)
We know the standard limits:
\( \text{Lim}_{x \to 0} \frac{e^x-1}{x} = \log_e e = 1 \)
\( \text{Lim}_{x \to 0} \frac{\sin x}{x} = 1 \)
So, the limit is \( 1 - 1 = 0 \). This illustrates how limits can be applied to individual terms of an expression if the individual limits exist.
(iv) \( \text{Lim}_{x \to 0} e^x \)
Since \( e^x \) is a continuous function, we can directly substitute \( x=0 \).
\( = e^0 = 1 \). For continuous functions, the limit at a point is simply the function's value at that point.
In simple words: For part (i), we used a trigonometry rule and broke the problem into two parts using standard limit rules. For part (ii), we did something similar, using a different trigonometry rule and then the same standard limit rules. For part (iii), we simply split the problem into two separate limits we already know. For part (iv), since it's a simple exponential function, we just put 0 in place of 'x'.

🎯 Exam Tip: Always look for ways to simplify trigonometric expressions using identities, and then try to rearrange the terms to match known limit formulas (like \( \frac{e^x-1}{x} \) or \( \frac{\sin x}{x} \)).

Question 3.
(i) \( \text{Lim}_{x \to 0} \frac{e^{ax}-e^{bx}}{x} \)
(ii) \( \text{Lim}_{x \to \frac{\pi}{2}} \frac{e^{\sin x}-1}{\sin x} \)
(iii) \( \text{Lim}_{x \to 2} \frac{e^x-e^2}{x-2} \)
(iv) \( \text{Lim}_{x \to 1} \frac{e^x-e}{x-1} \)
Answer:
(i) \( \text{Lim}_{x \to 0} \frac{e^{ax}-e^{bx}}{x} \)
We can rewrite the numerator by subtracting and adding 1:
\( = \text{Lim}_{x \to 0} \frac{(e^{ax}-1) - (e^{bx}-1)}{x} \)
Now, split the fraction into two parts:
\( = \text{Lim}_{x \to 0} \frac{e^{ax}-1}{x} - \text{Lim}_{x \to 0} \frac{e^{bx}-1}{x} \)
Using the standard limit \( \text{Lim}_{x \to 0} \frac{e^{kx}-1}{x} = k \log_e e = k \).
So, the limit is \( a - b \). This method of adding and subtracting a term is a common technique to convert expressions into known limit forms.
(ii) \( \text{Lim}_{x \to \frac{\pi}{2}} \frac{e^{\sin x}-1}{\sin x} \)
Let \( y = \sin x \). As \( x \to \frac{\pi}{2} \), \( y \to \sin(\frac{\pi}{2}) = 1 \).
The limit becomes \( \text{Lim}_{y \to 1} \frac{e^y-1}{y} \).
Since the function \( f(y) = \frac{e^y-1}{y} \) is continuous at \( y=1 \), we can directly substitute \( y=1 \):
\( = \frac{e^1-1}{1} = e-1 \). This shows how a change of variable can simplify a limit problem, especially when dealing with trigonometric functions.
(iii) \( \text{Lim}_{x \to 2} \frac{e^x-e^2}{x-2} \)
Let \( x = 2+h \). As \( x \to 2 \), \( h \to 0 \).
The expression becomes \( \text{Lim}_{h \to 0} \frac{e^{2+h}-e^2}{(2+h)-2} \)
\( = \text{Lim}_{h \to 0} \frac{e^2 e^h - e^2}{h} \)
Factor out \( e^2 \):
\( = e^2 \text{Lim}_{h \to 0} \frac{e^h-1}{h} \)
Using the standard limit \( \text{Lim}_{h \to 0} \frac{e^h-1}{h} = \log_e e = 1 \).
So, the limit is \( e^2 \cdot 1 = e^2 \). This limit is a direct application of the definition of the derivative for \( e^x \) at \( x=2 \).
(iv) \( \text{Lim}_{x \to 1} \frac{e^x-e}{x-1} \)
Let \( x = 1+h \). As \( x \to 1 \), \( h \to 0 \).
The expression becomes \( \text{Lim}_{h \to 0} \frac{e^{1+h}-e}{(1+h)-1} \)
\( = \text{Lim}_{h \to 0} \frac{e^1 e^h - e}{h} \)
Factor out \( e \):
\( = e \text{Lim}_{h \to 0} \frac{e^h-1}{h} \)
Using the standard limit \( \text{Lim}_{h \to 0} \frac{e^h-1}{h} = \log_e e = 1 \).
So, the limit is \( e \cdot 1 = e \). This is another example of finding the derivative of \( e^x \) at a specific point, which is \( e^x \) itself.
In simple words: For part (i), we changed the top part of the fraction by adding and subtracting '1' to use a known limit rule. For part (ii), we replaced 'sin x' with 'y' to make it a simpler known form. For parts (iii) and (iv), we substituted 'x' with 'constant + h' to make the problem look like a standard rule involving 'e'.

🎯 Exam Tip: When evaluating limits of the form \( \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \), recognize it as the definition of the derivative \( f'(a) \). This can often simplify the solution process.

Question 4. \( \text{Lim}_{x \to 1} \frac{e^{-x}-e^{-1}}{x-1} \)
Answer:
To evaluate this limit, we substitute \( x = 1+h \). As \( x \to 1 \), \( h \to 0 \).
The expression becomes \( \text{Lim}_{h \to 0} \frac{e^{-(1+h)}-e^{-1}}{(1+h)-1} \)
\( = \text{Lim}_{h \to 0} \frac{e^{-1}e^{-h}-e^{-1}}{h} \)
Factor out \( e^{-1} \):
\( = e^{-1} \text{Lim}_{h \to 0} \frac{e^{-h}-1}{h} \)
We can rewrite \( \frac{e^{-h}-1}{h} \) as \( -1 \cdot \frac{e^{-h}-1}{-h} \).
Let \( k = -h \). As \( h \to 0 \), \( k \to 0 \).
\( = e^{-1} \cdot (-1) \text{Lim}_{k \to 0} \frac{e^k-1}{k} \)
Using the standard limit \( \text{Lim}_{k \to 0} \frac{e^k-1}{k} = \log_e e = 1 \).
So, the limit is \( e^{-1} \cdot (-1) \cdot 1 = -e^{-1} = -\frac{1}{e} \). This problem also represents the derivative of \( e^{-x} \) at \( x=1 \).
In simple words: We changed 'x' to '1+h' to make the limit go towards 0. Then, we pulled out 'e to the power of -1' and rearranged the remaining part to use a known limit rule. The final answer is negative 'e to the power of -1'.

🎯 Exam Tip: When dealing with \( e^{-x} \) in limit problems, remember that \( e^{-x}-1 \) can be related to \( \frac{e^k-1}{k} \) by adjusting the sign and denominator carefully.

Question 5. \( \text{Lim}_{x \to 0} \frac{3^x-1}{\sqrt{1+x}-1} \)
Answer:
To evaluate this limit, we multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{1+x}+1 \).
\( = \text{Lim}_{x \to 0} \frac{(3^x-1)(\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)} \)
\( = \text{Lim}_{x \to 0} \frac{(3^x-1)(\sqrt{1+x}+1)}{(1+x)-1} \)
\( = \text{Lim}_{x \to 0} \frac{(3^x-1)(\sqrt{1+x}+1)}{x} \)
We can split this into a product of two limits:
\( = \text{Lim}_{x \to 0} \frac{3^x-1}{x} \cdot \text{Lim}_{x \to 0} (\sqrt{1+x}+1) \)
We know the standard limit \( \text{Lim}_{x \to 0} \frac{a^x-1}{x} = \log_e a \). So, the first part is \( \log_e 3 \).
For the second part, substitute \( x=0 \): \( \sqrt{1+0}+1 = 1+1 = 2 \).
So, the limit is \( \log_e 3 \cdot 2 = 2 \log_e 3 \). Multiplying by the conjugate is a standard technique used to rationalize denominators involving square roots in limits.
In simple words: We multiplied the top and bottom of the fraction by the 'plus' version of the bottom part. This helped to get rid of the square root downstairs. Then, we split the problem into two easier parts: one with a known rule for powers, and another where we could just put '0' for 'x'.

🎯 Exam Tip: When you see square roots in the denominator in a limit problem, consider multiplying by the conjugate to simplify the expression and eliminate the radical from the denominator.

Question 6. \( \text{Lim}_{x \to 1} x^{\frac{1}{x-1}} \)
Answer:
This limit is of the indeterminate form \( 1^\infty \).
To solve it, we use the substitution \( x = 1+h \). As \( x \to 1 \), \( h \to 0 \).
The expression becomes \( \text{Lim}_{h \to 0} (1+h)^{\frac{1}{(1+h)-1}} \)
\( = \text{Lim}_{h \to 0} (1+h)^{\frac{1}{h}} \).
This is a fundamental limit definition, which is equal to \( e \). The number 'e' appears naturally in many areas of mathematics and science, including compound interest and exponential growth.
In simple words: We changed 'x' to '1+h' to make 'h' go to zero. The expression then became a special form that directly equals the mathematical constant 'e'.

🎯 Exam Tip: Recognize the fundamental limit \( \lim_{h \to 0} (1+h)^{\frac{1}{h}} = e \). Many limits of the form \( 1^\infty \) can be transformed into this basic form through substitution and algebraic manipulation.

Question 7. \( \text{Lim}_{x \to 0} (1 + \sin x)^{\cot x} \)
Answer:
This limit is of the indeterminate form \( 1^\infty \).
We can rewrite \( \cot x \) as \( \frac{\cos x}{\sin x} \).
So, the expression becomes \( \text{Lim}_{x \to 0} (1 + \sin x)^{\frac{\cos x}{\sin x}} \).
We can further rewrite the exponent as \( \frac{1}{\sin x} \cdot \cos x \).
\( = \text{Lim}_{x \to 0} \left( (1 + \sin x)^{\frac{1}{\sin x}} \right)^{\cos x} \)
We use the standard limit form \( \text{Lim}_{f(x) \to 0} (1 + f(x))^{\frac{1}{f(x)}} = e \).
As \( x \to 0 \), \( \sin x \to 0 \). So, the inner part \( \text{Lim}_{x \to 0} (1 + \sin x)^{\frac{1}{\sin x}} = e \).
For the outer exponent, \( \text{Lim}_{x \to 0} \cos x = \cos 0 = 1 \).
Therefore, the overall limit is \( e^1 = e \). This technique transforms the limit into a recognizable standard form involving the constant 'e'.
In simple words: This problem looked like '1 to the power of infinity'. We changed 'cot x' to 'cos x divided by sin x'. Then, we used a special rule for limits that are '1 plus something, all raised to the power of 1 divided by that same something', which gives 'e'. The rest of the power just became 1.

🎯 Exam Tip: For limits of the form \( 1^\infty \), try to manipulate the expression to match \( \lim_{f(x) \to 0} (1 + f(x))^{\frac{1}{f(x)}} = e \). This often involves splitting the exponent into two parts.

Question 8. \( \text{Lim}_{x \to 0} \frac{8^x-2^x}{x} \)
Answer:
To find this limit, we rewrite the numerator by subtracting and adding 1:
\( = \text{Lim}_{x \to 0} \frac{(8^x-1) - (2^x-1)}{x} \)
Now, we split the fraction into two parts:
\( = \text{Lim}_{x \to 0} \frac{8^x-1}{x} - \text{Lim}_{x \to 0} \frac{2^x-1}{x} \)
Using the standard limit formula \( \text{Lim}_{x \to 0} \frac{a^x-1}{x} = \log_e a \).
So, the limit is \( \log_e 8 - \log_e 2 \).
Using logarithm properties, \( \log_e A - \log_e B = \log_e \left(\frac{A}{B}\right) \):
\( = \log_e \left(\frac{8}{2}\right) = \log_e 4 \). This problem demonstrates how the properties of logarithms can be used to simplify the final answer of a limit problem.
In simple words: We made the top part of the fraction work with a known rule by adding and subtracting '1'. Then, we split it into two simple limits. After finding the natural logarithm for each, we used a logarithm rule to combine them into one answer.

🎯 Exam Tip: Remember that \( \lim_{x \to 0} \frac{a^x-1}{x} = \log_e a \) is a powerful formula. If you have multiple exponential terms in the numerator, try to separate them using this form.

Question 9. \( \text{Lim}_{x \to 0} \frac{a^x-b^x}{\sin x} \)
Answer:
To find this limit, we first rewrite the numerator by subtracting and adding 1:
\( = \text{Lim}_{x \to 0} \frac{(a^x-1) - (b^x-1)}{\sin x} \)
Now, we divide both the numerator and denominator by \( x \):
\( = \text{Lim}_{x \to 0} \frac{\frac{a^x-1}{x} - \frac{b^x-1}{x}}{\frac{\sin x}{x}} \)
We can evaluate the limit of each part separately:
Numerator: \( \text{Lim}_{x \to 0} \frac{a^x-1}{x} - \text{Lim}_{x \to 0} \frac{b^x-1}{x} = \log_e a - \log_e b \).
Denominator: \( \text{Lim}_{x \to 0} \frac{\sin x}{x} = 1 \).
So, the overall limit is \( \frac{\log_e a - \log_e b}{1} = \log_e a - \log_e b \).
Using logarithm properties, \( \log_e a - \log_e b = \log_e \left(\frac{a}{b}\right) \). This problem combines both exponential and trigonometric limit forms, requiring careful application of standard formulas.
In simple words: We changed the top part by adding and subtracting '1'. Then, we divided both the top and bottom of the fraction by 'x'. This allowed us to use known rules for powers and 'sin x divided by x'. Finally, we used a logarithm rule to simplify the answer.

🎯 Exam Tip: When an expression contains both exponential and trigonometric terms, consider dividing by 'x' (or \( x^n \)) to reduce it to standard limit forms like \( \frac{a^x-1}{x} \) and \( \frac{\sin x}{x} \).

Question 10. \( \text{Lim}_{x \to 0} \frac{a^{\sin x}-1}{\sin x} \)
Answer:
To evaluate this limit, we use a substitution.
Let \( t = \sin x \). As \( x \to 0 \), \( \sin x \to 0 \), so \( t \to 0 \).
The expression then becomes \( \text{Lim}_{t \to 0} \frac{a^t-1}{t} \).
This is a standard limit formula, which is equal to \( \log_e a \). Substitution is a powerful technique for transforming limits into familiar forms.
In simple words: We replaced 'sin x' with 't'. As 'x' went to zero, 't' also went to zero. This made the problem look like a simple, known limit rule for powers, which gives the natural logarithm of 'a'.

🎯 Exam Tip: When you see a complex function inside a standard limit form (e.g., \( \sin x \) inside \( \frac{a^k-1}{k} \)), consider using a substitution to simplify the expression and make the standard form more apparent.

Question 11. \( \text{Lim}_{x \to 0} \frac{3^{2x}-2^{3x}}{x} \)
Answer:
To find this limit, we rewrite the numerator by subtracting and adding 1:
\( = \text{Lim}_{x \to 0} \frac{(3^{2x}-1) - (2^{3x}-1)}{x} \)
Now, we split the fraction into two parts:
\( = \text{Lim}_{x \to 0} \frac{3^{2x}-1}{x} - \text{Lim}_{x \to 0} \frac{2^{3x}-1}{x} \)
For the first part, \( \text{Lim}_{x \to 0} \frac{3^{2x}-1}{x} \): we multiply and divide by 2 to get \( 2 \text{Lim}_{x \to 0} \frac{3^{2x}-1}{2x} = 2 \log_e 3 \).
For the second part, \( \text{Lim}_{x \to 0} \frac{2^{3x}-1}{x} \): we multiply and divide by 3 to get \( 3 \text{Lim}_{x \to 0} \frac{2^{3x}-1}{3x} = 3 \log_e 2 \).
So, the overall limit is \( 2 \log_e 3 - 3 \log_e 2 \).
Using logarithm properties, \( k \log_e a = \log_e (a^k) \):
\( = \log_e (3^2) - \log_e (2^3) = \log_e 9 - \log_e 8 \).
Then, using \( \log_e A - \log_e B = \log_e \left(\frac{A}{B}\right) \):
\( = \log_e \left(\frac{9}{8}\right) \). This problem highlights the importance of matching the variable in the exponent with the variable in the denominator when using the limit formula for \( a^x \).
In simple words: We changed the top part of the fraction by adding and subtracting '1'. Then, we split the problem into two limits. For each limit, we adjusted the fraction by multiplying and dividing by a number so it fit the known rule for powers. Finally, we used logarithm rules to simplify the answer.

🎯 Exam Tip: For expressions like \( \frac{a^{kx}-1}{x} \), always multiply the numerator and denominator by 'k' to get \( k \frac{a^{kx}-1}{kx} \), allowing you to apply the standard formula \( k \log_e a \).

Question 12. \( \text{Lim}_{x \to 1} \frac{x-1}{\log_e x} \)
Answer:
To evaluate this limit, we use the substitution \( x = 1+h \). As \( x \to 1 \), \( h \to 0 \).
The expression becomes \( \text{Lim}_{h \to 0} \frac{(1+h)-1}{\log_e (1+h)} \)
\( = \text{Lim}_{h \to 0} \frac{h}{\log_e (1+h)} \).
We can rewrite this as \( \frac{1}{\text{Lim}_{h \to 0} \frac{\log_e (1+h)}{h}} \).
The denominator is a standard limit, \( \text{Lim}_{h \to 0} \frac{\log_e (1+h)}{h} = \log_e e = 1 \).
Therefore, the overall limit is \( \frac{1}{1} = 1 \). This limit is directly related to the derivative of \( \log_e x \) at \( x=1 \).
In simple words: We changed 'x' to '1+h' to make the limit go towards 0. The problem then became 'h divided by log(1+h)'. We know that 'log(1+h) divided by h' goes to 1, so the inverse also goes to 1.

🎯 Exam Tip: Remember the reciprocal relationship: if \( \lim_{x \to 0} \frac{f(x)}{g(x)} = L \) and \( L \ne 0 \), then \( \lim_{x \to 0} \frac{g(x)}{f(x)} = \frac{1}{L} \). This is useful when the standard limit form appears inverted.

Question 13. \( \text{Lim}_{x \to 0} \frac{e^x+e^{-x}-2}{x^2} \)
Answer:
To find this limit, we first combine the terms in the numerator by taking a common denominator of \( e^x \):
\( = \text{Lim}_{x \to 0} \frac{e^x(e^x)+e^x(e^{-x})-2e^x}{x^2 e^x} \)
\( = \text{Lim}_{x \to 0} \frac{e^{2x}+1-2e^x}{x^2 e^x} \)
The numerator \( e^{2x}-2e^x+1 \) is a perfect square: \( (e^x-1)^2 \).
So, the expression becomes \( \text{Lim}_{x \to 0} \frac{(e^x-1)^2}{x^2 e^x} \).
We can rearrange this as a product of limits:
\( = \text{Lim}_{x \to 0} \left( \frac{e^x-1}{x} \right)^2 \cdot \text{Lim}_{x \to 0} \frac{1}{e^x} \)
We know the standard limit \( \text{Lim}_{x \to 0} \frac{e^x-1}{x} = \log_e e = 1 \). So, the first part is \( (1)^2 = 1 \).
For the second part, substitute \( x=0 \): \( \frac{1}{e^0} = \frac{1}{1} = 1 \).
Therefore, the overall limit is \( 1 \cdot 1 = 1 \). Recognizing algebraic patterns like perfect squares can greatly simplify limit calculations.
In simple words: We made the top part of the fraction simpler by putting everything over 'e to the power of x', then we saw it was a perfect square. After that, we broke the problem into two limits that we already knew how to solve, which both equaled 1.

🎯 Exam Tip: Look for opportunities to simplify algebraic expressions in the numerator or denominator (like factoring or combining terms) before applying limit formulas. Perfect squares and differences of squares are common patterns.

Question 14. \( \text{Lim}_{x \to 5} \frac{\log x-\log 5}{x-5} \)
Answer:
To find this limit, we use the substitution \( x = 5+h \). As \( x \to 5 \), \( h \to 0 \).
The expression becomes \( \text{Lim}_{h \to 0} \frac{\log(5+h)-\log 5}{(5+h)-5} \)
\( = \text{Lim}_{h \to 0} \frac{\log\left(\frac{5+h}{5}\right)}{h} \)
\( = \text{Lim}_{h \to 0} \frac{\log\left(1+\frac{h}{5}\right)}{h} \).
To use the standard limit \( \text{Lim}_{y \to 0} \frac{\log(1+y)}{y} = 1 \), we multiply and divide the term inside the limit by \( \frac{1}{5} \):
\( = \text{Lim}_{h \to 0} \frac{1}{5} \cdot \frac{\log\left(1+\frac{h}{5}\right)}{\frac{h}{5}} \)
As \( h \to 0 \), \( \frac{h}{5} \to 0 \). So, the part \( \text{Lim}_{h \to 0} \frac{\log\left(1+\frac{h}{5}\right)}{\frac{h}{5}} = 1 \).
Therefore, the overall limit is \( \frac{1}{5} \cdot 1 = \frac{1}{5} \). This is a direct application of the derivative of \( \log_e x \) at \( x=5 \).
In simple words: We changed 'x' to '5+h' to make the limit go towards 0. Then, we used a logarithm rule to combine the two 'log' terms at the top. After that, we adjusted the fraction so it fit a known limit rule for logarithms, which gave us the answer.

🎯 Exam Tip: When dealing with logarithmic limits like \( \lim_{x \to a} \frac{\log x - \log a}{x-a} \), recall the derivative definition of \( \log x \) at 'a'. It always simplifies to \( \frac{1}{a} \).

Question 15. \( \text{Lim}_{x \to 0} \frac{e^x-1}{\sqrt{1-\cos x}} \)
Answer:
To evaluate this limit, we use the trigonometric identity \( 1-\cos x = 2\sin^2 \frac{x}{2} \).
The expression becomes \( \text{Lim}_{x \to 0} \frac{e^x-1}{\sqrt{2\sin^2 \frac{x}{2}}} \)
\( = \text{Lim}_{x \to 0} \frac{e^x-1}{\sqrt{2} \left| \sin \frac{x}{2} \right|} \).
For \( x \to 0^+ \) (from the right side), \( \sin \frac{x}{2} > 0 \), so \( \left| \sin \frac{x}{2} \right| = \sin \frac{x}{2} \).
\( = \text{Lim}_{x \to 0^+} \frac{e^x-1}{\sqrt{2} \sin \frac{x}{2}} \)
We multiply and divide by \( x \) in the numerator and denominator:
\( = \text{Lim}_{x \to 0^+} \frac{\frac{e^x-1}{x}}{\frac{\sqrt{2} \sin \frac{x}{2}}{x}} \)
Numerator: \( \text{Lim}_{x \to 0^+} \frac{e^x-1}{x} = 1 \).
Denominator: \( \text{Lim}_{x \to 0^+} \frac{\sqrt{2} \sin \frac{x}{2}}{x} = \text{Lim}_{x \to 0^+} \frac{\sqrt{2}}{2} \frac{\sin \frac{x}{2}}{\frac{x}{2}} = \frac{1}{\sqrt{2}} \cdot 1 = \frac{1}{\sqrt{2}} \).
So, the right-hand limit is \( \frac{1}{1/\sqrt{2}} = \sqrt{2} \).
For \( x \to 0^- \) (from the left side), \( \sin \frac{x}{2} < 0 \), so \( \left| \sin \frac{x}{2} \right| = -\sin \frac{x}{2} \).
The left-hand limit will be \( \frac{1}{-1/\sqrt{2}} = -\sqrt{2} \).
Since the left-hand limit (\( -\sqrt{2} \)) and the right-hand limit (\( \sqrt{2} \)) are not equal, the limit does not exist. However, if the question intends for the principal square root in the positive domain, then the answer is \( \sqrt{2} \). The provided solution implies the positive branch only.
Following the provided solution's calculation path (assuming \( \sqrt{\sin^2 \frac{x}{2}} = \sin \frac{x}{2} \) for \( x \to 0 \)):
\( = \text{Lim}_{x \to 0} \frac{e^x-1}{x} \cdot \frac{x}{\sqrt{2}\sin\frac{x}{2}} \)
\( = 1 \cdot \text{Lim}_{x \to 0} \frac{1}{\sqrt{2}} \cdot \frac{x}{\sin\frac{x}{2}} \)
\( = \frac{1}{\sqrt{2}} \cdot \text{Lim}_{x \to 0} \frac{2 \cdot \frac{x}{2}}{\sin\frac{x}{2}} = \frac{1}{\sqrt{2}} \cdot 2 \cdot 1 = \sqrt{2} \). When dealing with square roots of squared terms like \( \sqrt{\sin^2 \theta} \), remember that it simplifies to \( |\sin \theta| \), which might require considering left and right-hand limits if \( \theta \to 0 \).
In simple words: First, we used a trigonometry rule to change the bottom part of the fraction. Then, we separated the problem into two parts that used known limit rules. One part dealt with 'e to the power of x', and the other with 'sin x over x'. We found that the limit from the right side gave positive root 2, and from the left side gave negative root 2, meaning the limit doesn't exist. The given solution only shows the positive case.

🎯 Exam Tip: Be cautious with square roots in limits, especially around points where the expression inside the root changes sign. \( \sqrt{f(x)^2} \) is \( |f(x)| \), which requires evaluating left-hand and right-hand limits if \( f(x) \to 0 \).

Question 16.
(i) \( \text{Lim}_{n \to \infty} \left(1 + \frac{2}{n}\right)^{2n} \)
(ii) \( \text{Lim}_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx} \)
(iii) \( \text{Lim}_{x \to 0} (1 + ax)^{\frac{b}{x}} \)
Answer:
(i) \( \text{Lim}_{n \to \infty} \left(1 + \frac{2}{n}\right)^{2n} \)
This limit is of the form \( 1^\infty \). We use the standard limit form \( \text{Lim}_{x \to \infty} \left(1 + \frac{a}{x}\right)^{kx} = e^{ak} \).
Here, \( a=2 \) and \( k=2 \).
So, the limit is \( e^{2 \cdot 2} = e^4 \). This limit is a direct application of a common formula involving the constant 'e' and powers tending to infinity.
(ii) \( \text{Lim}_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx} \)
This limit is also of the form \( 1^\infty \). Let \( y = a+bx \). As \( x \to \infty \), \( y \to \infty \).
The expression can be rewritten as \( \text{Lim}_{x \to \infty} \left( \left(1 + \frac{1}{a+bx}\right)^{a+bx} \right)^{\frac{c+dx}{a+bx}} \)
The inner part, \( \text{Lim}_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{a+bx} \), is of the form \( \text{Lim}_{y \to \infty} \left(1 + \frac{1}{y}\right)^y = e \).
For the exponent, \( \text{Lim}_{x \to \infty} \frac{c+dx}{a+bx} \). Divide numerator and denominator by \( x \):
\( = \text{Lim}_{x \to \infty} \frac{\frac{c}{x}+d}{\frac{a}{x}+b} = \frac{0+d}{0+b} = \frac{d}{b} \).
Therefore, the overall limit is \( e^{\frac{d}{b}} \). When the base and exponent both involve variables, splitting the expression and evaluating each limit separately is often helpful.
(iii) \( \text{Lim}_{x \to 0} (1 + ax)^{\frac{b}{x}} \)
This limit is of the form \( 1^\infty \). We use the standard limit form \( \text{Lim}_{x \to 0} (1 + kx)^{\frac{1}{x}} = e^k \).
We can rewrite the exponent as \( \frac{b}{x} = \frac{1}{ax} \cdot ab \).
So, the expression becomes \( \text{Lim}_{x \to 0} \left( (1 + ax)^{\frac{1}{ax}} \right)^{ab} \)
The inner part, \( \text{Lim}_{x \to 0} (1 + ax)^{\frac{1}{ax}} \), is of the form \( \text{Lim}_{y \to 0} (1+y)^{\frac{1}{y}} = e \).
Therefore, the overall limit is \( e^{ab} \). For limits involving powers, transforming the expression to match the form \( (1+f(x))^{1/f(x)} \) is a common strategy.
In simple words: All these problems are '1 to the power of infinity' type limits. For each, we reshaped the expression to match a special rule where the answer is 'e' raised to some power. We found this power by looking at the numbers and variables in the base and exponent.

🎯 Exam Tip: Mastering the three fundamental forms of limits involving 'e' ( \( \lim_{x \to \infty} (1 + \frac{1}{x})^x \), \( \lim_{x \to \infty} (1 + \frac{a}{x})^{bx} \), and \( \lim_{x \to 0} (1+ax)^{1/x} \)) is crucial for solving such problems quickly and accurately.

Question 17.
(i) \( \text{Lim}_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x+4} \)
(ii) \( \text{Lim}_{x \to \infty} \left(\frac{x-1}{x+1}\right)^x \)
(iii) \( \text{Lim}_{x \to 0} \left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{x^2}} \)
Answer:
(i) \( \text{Lim}_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x+4} \)
This limit is of the form \( 1^\infty \). First, rewrite the base:
\( \frac{x+6}{x+1} = \frac{(x+1)+5}{x+1} = 1 + \frac{5}{x+1} \).
So, the expression becomes \( \text{Lim}_{x \to \infty} \left(1 + \frac{5}{x+1}\right)^{x+4} \).
We use the standard limit form \( \text{Lim}_{y \to \infty} \left(1 + \frac{a}{y}\right)^{ky} = e^{ak} \).
Here, \( a=5 \). Let \( y = x+1 \). Then as \( x \to \infty \), \( y \to \infty \). The exponent \( x+4 = (x+1)+3 = y+3 \).
So, the limit is \( \text{Lim}_{y \to \infty} \left(1 + \frac{5}{y}\right)^{y+3} = \text{Lim}_{y \to \infty} \left(1 + \frac{5}{y}\right)^y \cdot \left(1 + \frac{5}{y}\right)^3 \).
The first part is \( e^5 \). The second part, as \( y \to \infty \), becomes \( (1+0)^3 = 1 \).
Therefore, the overall limit is \( e^5 \cdot 1 = e^5 \). For limits of this type, manipulating the base to be of the form \( 1 + \frac{a}{y} \) is the key step.
(ii) \( \text{Lim}_{x \to \infty} \left(\frac{x-1}{x+1}\right)^x \)
This limit is also of the form \( 1^\infty \). First, rewrite the base:
\( \frac{x-1}{x+1} = \frac{(x+1)-2}{x+1} = 1 - \frac{2}{x+1} \).
So, the expression becomes \( \text{Lim}_{x \to \infty} \left(1 - \frac{2}{x+1}\right)^x \).
Let \( y = x+1 \). Then as \( x \to \infty \), \( y \to \infty \). The exponent \( x = y-1 \).
The limit is \( \text{Lim}_{y \to \infty} \left(1 - \frac{2}{y}\right)^{y-1} = \text{Lim}_{y \to \infty} \left(1 - \frac{2}{y}\right)^y \cdot \left(1 - \frac{2}{y}\right)^{-1} \).
The first part is \( e^{-2} \). The second part, as \( y \to \infty \), becomes \( (1-0)^{-1} = 1 \).
Therefore, the overall limit is \( e^{-2} \cdot 1 = e^{-2} \). When using the standard limit \( \left(1 + \frac{a}{y}\right)^y \), ensure that the 'y' in the denominator and the exponent are identical.
(iii) \( \text{Lim}_{x \to 0} \left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{x^2}} \)
This limit is also of the form \( 1^\infty \). First, rewrite the base:
\( \frac{1+5x^2}{1+3x^2} = \frac{(1+3x^2)+2x^2}{1+3x^2} = 1 + \frac{2x^2}{1+3x^2} \).
So, the expression becomes \( \text{Lim}_{x \to 0} \left(1 + \frac{2x^2}{1+3x^2}\right)^{\frac{1}{x^2}} \).
Let \( f(x) = \frac{2x^2}{1+3x^2} \). As \( x \to 0 \), \( f(x) \to 0 \). We want to use \( \text{Lim}_{y \to 0} (1+y)^{\frac{1}{y}} = e \).
The exponent is \( \frac{1}{x^2} \). We can write this as \( \frac{1+3x^2}{2x^2} \cdot \frac{2}{1+3x^2} \).
So, the limit becomes \( \text{Lim}_{x \to 0} \left( \left(1 + \frac{2x^2}{1+3x^2}\right)^{\frac{1+3x^2}{2x^2}} \right)^{\frac{2}{1+3x^2}} \).
The inner part approaches \( e \). The outer exponent \( \text{Lim}_{x \to 0} \frac{2}{1+3x^2} = \frac{2}{1+0} = 2 \).
Therefore, the overall limit is \( e^2 \). This method of creating the reciprocal of the added term in the base as part of the exponent is essential for these types of limits.
In simple words: All these limits were '1 to the power of infinity'. For each, we first changed the base of the power to look like '1 plus a small fraction'. Then, we adjusted the exponent to fit a special rule that gives 'e' raised to some power.

🎯 Exam Tip: When the base is a fraction, simplify it to the form \( 1 \pm \frac{A}{B} \). Then, identify the `A/B` part to use as the `y` for the standard `(1+y)^(1/y)` limit form.

Question 18.
(i) \( \text{Lim}_{x \to 0} \frac{e^{ax}-1}{\sin x} \)
(ii) \( \text{Lim}_{x \to 0} \frac{e^{x^2}-1}{\sin x^2} \)
(iii) \( \text{Lim}_{x \to 0} \frac{3^{2x}-1}{2^{3x}-1} \)
(iv) \( \text{Lim}_{x \to 1} \frac{\sin (e^{x-1}-1)}{\log x} \)
Answer:
(i) \( \text{Lim}_{x \to 0} \frac{e^{ax}-1}{\sin x} \)
To evaluate this limit, we divide both the numerator and denominator by \( x \):
\( = \text{Lim}_{x \to 0} \frac{\frac{e^{ax}-1}{x}}{\frac{\sin x}{x}} \)
We can evaluate the limit of each part separately:
Numerator: \( \text{Lim}_{x \to 0} \frac{e^{ax}-1}{x} = a \log_e e = a \).
Denominator: \( \text{Lim}_{x \to 0} \frac{\sin x}{x} = 1 \).
Therefore, the overall limit is \( \frac{a}{1} = a \). This problem uses a common strategy of dividing by 'x' to transform the expression into standard limit forms.
(ii) \( \text{Lim}_{x \to 0} \frac{e^{x^2}-1}{\sin x^2} \)
To evaluate this limit, we divide both the numerator and denominator by \( x^2 \):
\( = \text{Lim}_{x \to 0} \frac{\frac{e^{x^2}-1}{x^2}}{\frac{\sin x^2}{x^2}} \)
Let \( y = x^2 \). As \( x \to 0 \), \( y \to 0 \).
Numerator: \( \text{Lim}_{y \to 0} \frac{e^y-1}{y} = \log_e e = 1 \).
Denominator: \( \text{Lim}_{y \to 0} \frac{\sin y}{y} = 1 \).
Therefore, the overall limit is \( \frac{1}{1} = 1 \). Using a substitution like \( y = x^2 \) makes it easier to recognize the standard limit forms.
(iii) \( \text{Lim}_{x \to 0} \frac{3^{2x}-1}{2^{3x}-1} \)
To evaluate this limit, we divide both the numerator and denominator by \( x \):
\( = \text{Lim}_{x \to 0} \frac{\frac{3^{2x}-1}{x}}{\frac{2^{3x}-1}{x}} \)
For the numerator, \( \text{Lim}_{x \to 0} \frac{3^{2x}-1}{x} = 2 \text{Lim}_{x \to 0} \frac{3^{2x}-1}{2x} = 2 \log_e 3 \).
For the denominator, \( \text{Lim}_{x \to 0} \frac{2^{3x}-1}{x} = 3 \text{Lim}_{x \to 0} \frac{2^{3x}-1}{3x} = 3 \log_e 2 \).
Therefore, the overall limit is \( \frac{2 \log_e 3}{3 \log_e 2} \).
Using logarithm properties, \( \frac{\log_e (3^2)}{\log_e (2^3)} = \frac{\log_e 9}{\log_e 8} \). Remember to adjust the constants in the denominator to match the exponent when using the standard limit for \( a^x \).
(iv) \( \text{Lim}_{x \to 1} \frac{\sin (e^{x-1}-1)}{\log x} \)
To evaluate this limit, we substitute \( x = 1+h \). As \( x \to 1 \), \( h \to 0 \).
The expression becomes \( \text{Lim}_{h \to 0} \frac{\sin (e^{(1+h)-1}-1)}{\log (1+h)} = \text{Lim}_{h \to 0} \frac{\sin (e^h-1)}{\log (1+h)} \).
We can multiply and divide by appropriate terms to use standard limit forms:
\( = \text{Lim}_{h \to 0} \left( \frac{\sin (e^h-1)}{e^h-1} \cdot \frac{e^h-1}{h} \cdot \frac{h}{\log (1+h)} \right) \)
Let's evaluate each part:
1. \( \text{Lim}_{h \to 0} \frac{\sin (e^h-1)}{e^h-1} \). Let \( y = e^h-1 \). As \( h \to 0 \), \( y \to 0 \). So, \( \text{Lim}_{y \to 0} \frac{\sin y}{y} = 1 \).
2. \( \text{Lim}_{h \to 0} \frac{e^h-1}{h} = \log_e e = 1 \).
3. \( \text{Lim}_{h \to 0} \frac{h}{\log (1+h)} = \frac{1}{\text{Lim}_{h \to 0} \frac{\log (1+h)}{h}} = \frac{1}{1} = 1 \).
Multiplying these results, the overall limit is \( 1 \times 1 \times 1 = 1 \). This limit elegantly combines substitutions, trigonometric limits, exponential limits, and logarithmic limits into a single problem.
In simple words: For the first three parts, we used the trick of dividing the top and bottom of the fraction by 'x' (or 'x squared') to make them fit known limit rules. For the fourth part, we changed 'x' to '1+h' and then broke the problem into three simple parts, each of which we know equals 1.

🎯 Exam Tip: For complex limits, try to use substitution to transform the variable and then creatively multiply/divide by terms to isolate standard limit forms. Ensure all individual limits exist before multiplying them.

Question 19. Evaluate: \( \text{Lim}_{x \to 0} \frac{\log (a + x) - \log (a - x)}{x}, a > 0 \)
Answer: We need to evaluate the given limit.
\[ \text{Lim}_{x \to 0} \frac{\log (a + x) - \log (a - x)}{x} \] First, we use the logarithm property \( \log M - \log N = \log \left( \frac{M}{N} \right) \): \[ = \text{Lim}_{x \to 0} \frac{1}{x} \log \left( \frac{a + x}{a - x} \right) \] Next, we rewrite the fraction inside the logarithm to fit the standard limit form \( \text{Lim}_{y \to 0} \frac{\log(1+y)}{y} = 1 \). We can write \( \frac{a+x}{a-x} = \frac{a-x+2x}{a-x} = 1 + \frac{2x}{a-x} \). \[ = \text{Lim}_{x \to 0} \frac{1}{x} \log \left( 1 + \frac{2x}{a - x} \right) \] To apply the standard limit, we need \( \frac{2x}{a-x} \) in the denominator. We multiply and divide by this term: \[ = \text{Lim}_{x \to 0} \frac{\log \left( 1 + \frac{2x}{a - x} \right)}{\frac{2x}{a - x}} \cdot \frac{2x}{a - x} \cdot \frac{1}{x} \] As \( x \to 0 \), the term \( \frac{2x}{a-x} \to 0 \). So, the first part of the expression simplifies to 1 based on the standard limit. \[ = 1 \cdot \text{Lim}_{x \to 0} \frac{2x}{x(a - x)} \] We can cancel \( x \) from the numerator and denominator: \[ = \text{Lim}_{x \to 0} \frac{2}{a - x} \] Now, substitute \( x = 0 \) into the expression: \[ = \frac{2}{a - 0} \] \[ = \frac{2}{a} \]In simple words: We start by combining the two log terms using a basic rule. Then, we change the fraction inside the log so it looks like "1 plus something". This lets us use a special limit rule. After simplifying and cancelling terms, we put \( x=0 \) to get the final answer.

🎯 Exam Tip: When evaluating limits with logarithmic functions, always try to transform the expression into the standard form \( \text{Lim}_{y \to 0} \frac{\log(1+y)}{y} = 1 \) or related forms. Algebraic manipulation is key here.