Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.5

Question 1. Evaluate the following limits, if necessary use L' Hôpital's Rule: \( \lim _{x \rightarrow 0} \frac { 1-\cos x }{ x^2 } \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow 0} \frac { 1-\cos x }{ x^2 } \) Substitute \( x = 0 \): \( \frac { 1-\cos(0) }{ 0^2 } = \frac { 1-1 }{ 0 } = \frac { 0 }{ 0 } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule (take derivatives of numerator and denominator): Numerator derivative: \( \frac{d}{dx}(1-\cos x) = \sin x \) Denominator derivative: \( \frac{d}{dx}(x^2) = 2x \)
\( \implies \lim _{x \rightarrow 0} \frac { \sin x }{ 2x } \) Substitute \( x = 0 \): \( \frac { \sin(0) }{ 2(0) } = \frac { 0 }{ 0 } \) This is still an indeterminate form, so apply L' Hôpital's Rule again.
\( \implies \) Apply L' Hôpital's Rule again: Numerator derivative: \( \frac{d}{dx}(\sin x) = \cos x \) Denominator derivative: \( \frac{d}{dx}(2x) = 2 \)
\( \implies \lim _{x \rightarrow 0} \frac { \cos x }{ 2 } \) Now, substitute \( x = 0 \): \( \frac { \cos(0) }{ 2 } = \frac { 1 }{ 2 } \) The limit is \( \frac{1}{2} \). L' Hôpital's Rule helps solve limits that are in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
In simple words: When you try to find the value of this math problem at \( x=0 \), you get \( \frac{0}{0} \), which doesn't give a clear answer. So, we use a special rule called L' Hôpital's Rule. This rule tells us to take the derivative of the top part and the bottom part of the fraction and try again. We did this twice, and the final answer we got was \( \frac{1}{2} \).

🎯 Exam Tip: Remember to check the form of the limit (e.g., \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)) before applying L' Hôpital's Rule. Sometimes you might need to apply it multiple times.

Question 2. \( \lim _{x \rightarrow \infty} \frac { 2x^2-3 }{ x^2-5x+3 } \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow \infty} \frac { 2x^2-3 }{ x^2-5x+3 } \) Substitute \( x = \infty \): \( \frac { 2(\infty)^2-3 }{ (\infty)^2-5(\infty)+3 } = \frac { \infty }{ \infty } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule (take derivatives of numerator and denominator): Numerator derivative: \( \frac{d}{dx}(2x^2-3) = 4x \) Denominator derivative: \( \frac{d}{dx}(x^2-5x+3) = 2x-5 \)
\( \implies \lim _{x \rightarrow \infty} \frac { 4x }{ 2x-5 } \) Substitute \( x = \infty \): \( \frac { 4(\infty) }{ 2(\infty)-5 } = \frac { \infty }{ \infty } \) This is still an indeterminate form, so apply L' Hôpital's Rule again.
\( \implies \) Apply L' Hôpital's Rule again: Numerator derivative: \( \frac{d}{dx}(4x) = 4 \) Denominator derivative: \( \frac{d}{dx}(2x-5) = 2 \)
\( \implies \lim _{x \rightarrow \infty} \frac { 4 }{ 2 } = 2 \) The limit is 2. This method is especially useful for rational functions where the degree of the numerator and denominator are the same.
In simple words: When we try to find the value of this problem as \( x \) becomes very, very big (goes to infinity), we get \( \frac{\infty}{\infty} \), which is unclear. So, we use L' Hôpital's Rule. We take the derivative of the top and bottom parts twice. After doing that, the answer we get is 2.

🎯 Exam Tip: For limits of rational functions as \( x \rightarrow \infty \), you can also divide the numerator and denominator by the highest power of \( x \) to simplify. This often provides a quicker path to the answer.

Question 3. \( \lim _{x \rightarrow \infty} \frac { x }{ \log x } \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow \infty} \frac { x }{ \log x } \) Substitute \( x = \infty \): \( \frac { \infty }{ \log(\infty) } = \frac { \infty }{ \infty } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule (take derivatives of numerator and denominator): Numerator derivative: \( \frac{d}{dx}(x) = 1 \) Denominator derivative: \( \frac{d}{dx}(\log x) = \frac{1}{x} \)
\( \implies \lim _{x \rightarrow \infty} \frac { 1 }{ \frac{1}{x} } \)
\( \implies \lim _{x \rightarrow \infty} x \) Substitute \( x = \infty \): \( \infty \) The limit is \( \infty \). This shows that \( x \) grows much faster than \( \log x \) as \( x \) approaches infinity.
In simple words: We want to find what happens to this fraction as \( x \) gets very large. Since we get \( \frac{\infty}{\infty} \), we use L' Hôpital's Rule. We take the derivative of the top and bottom of the fraction. After doing this, the problem simplifies to just \( \lim _{x \rightarrow \infty} x \), which means the answer is also very large, or infinity.

🎯 Exam Tip: Remember that \( \log x \) (natural logarithm) grows much slower than any positive power of \( x \), which means \( x/\log x \) will always tend towards infinity.

Question 4. \( \lim _{x \rightarrow \frac{\pi}{2}} \frac { \sec x }{ \tan x } \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow \frac{\pi}{2}} \frac { \sec x }{ \tan x } \) Substitute \( x = \frac{\pi}{2} \): \( \sec(\frac{\pi}{2}) \) is undefined (approaches \( \infty \) or \( -\infty \)). \( \tan(\frac{\pi}{2}) \) is undefined (approaches \( \infty \) or \( -\infty \)). So, this is an \( \frac{\infty}{\infty} \) indeterminate form. We can simplify the expression first.
\( \implies \) Rewrite \( \sec x \) and \( \tan x \) in terms of \( \sin x \) and \( \cos x \): \( \frac { \sec x }{ \tan x } = \frac { \frac{1}{\cos x} }{ \frac{\sin x}{\cos x} } \)
\( \implies = \frac { 1 }{ \cos x } \cdot \frac { \cos x }{ \sin x } = \frac { 1 }{ \sin x } \)
\( \implies \) Now, evaluate the limit of the simplified expression: \( \lim _{x \rightarrow \frac{\pi}{2}} \frac { 1 }{ \sin x } \) Substitute \( x = \frac{\pi}{2} \): \( \frac { 1 }{ \sin(\frac{\pi}{2}) } = \frac { 1 }{ 1 } = 1 \) The limit is 1. Simplifying trigonometric expressions before evaluating the limit can often make the problem much easier to solve.
In simple words: We need to find the value of this fraction as \( x \) gets close to \( \frac{\pi}{2} \). At \( \frac{\pi}{2} \), both the top and bottom parts of the fraction become infinity, which is not a clear answer. So, we first change the `sec x` and `tan x` into `sin x` and `cos x`. After simplifying, the fraction becomes `1/sin x`. Now, putting \( x = \frac{\pi}{2} \) gives us `1/1`, which is 1.

🎯 Exam Tip: For limits involving trigonometric functions, always try to simplify the expression using identities before applying L' Hôpital's Rule or direct substitution. This can prevent unnecessary complex calculations.

Question 5. \( \lim _{x \rightarrow \infty} e^{-x}\sqrt{x} \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow \infty} e^{-x}\sqrt{x} \) Substitute \( x = \infty \): \( e^{-\infty}\sqrt{\infty} = 0 \cdot \infty \) This is an indeterminate form. We need to rewrite the expression to apply L' Hôpital's Rule.
\( \implies \) Rewrite the expression as a fraction: \( e^{-x}\sqrt{x} = \frac{\sqrt{x}}{e^x} \)
\( \implies \) Now, apply the limit to the rewritten expression: \( \lim _{x \rightarrow \infty} \frac{\sqrt{x}}{e^x} \) Substitute \( x = \infty \): \( \frac{\sqrt{\infty}}{e^{\infty}} = \frac{\infty}{\infty} \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule: Numerator derivative: \( \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \) Denominator derivative: \( \frac{d}{dx}(e^x) = e^x \)
\( \implies \lim _{x \rightarrow \infty} \frac{\frac{1}{2\sqrt{x}}}{e^x} \)
\( \implies \lim _{x \rightarrow \infty} \frac{1}{2\sqrt{x}e^x} \) Substitute \( x = \infty \): \( \frac{1}{2\sqrt{\infty}e^{\infty}} = \frac{1}{\infty} = 0 \) The limit is 0. This demonstrates that exponential functions grow much faster than polynomial or root functions.
In simple words: We are looking for the value of this problem as \( x \) becomes very large. When we put in infinity, we get `0 * infinity`, which is unclear. So, we rewrite the problem as a fraction. Then, we use L' Hôpital's Rule by taking the derivative of the top and bottom. After doing this, we see that the bottom part grows much faster, making the whole fraction go to 0.

🎯 Exam Tip: When faced with \( 0 \cdot \infty \) or \( \infty - \infty \) indeterminate forms, always try to convert them into \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form using algebraic manipulation before applying L' Hôpital's Rule.

Question 6. \( \lim _{x \rightarrow 0} \left( \frac { 1 }{ \sin x } – \frac { 1 }{ x } \right) \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow 0} \left( \frac { 1 }{ \sin x } – \frac { 1 }{ x } \right) \) Substitute \( x = 0 \): \( \frac { 1 }{ \sin(0) } – \frac { 1 }{ 0 } = \frac { 1 }{ 0 } – \frac { 1 }{ 0 } = \infty - \infty \) This is an indeterminate form. We need to combine the fractions to apply L' Hôpital's Rule.
\( \implies \) Combine the fractions: \( \frac { 1 }{ \sin x } – \frac { 1 }{ x } = \frac { x - \sin x }{ x \sin x } \)
\( \implies \) Now, apply the limit to the combined expression: \( \lim _{x \rightarrow 0} \frac { x - \sin x }{ x \sin x } \) Substitute \( x = 0 \): \( \frac { 0 - \sin(0) }{ 0 \cdot \sin(0) } = \frac { 0 - 0 }{ 0 \cdot 0 } = \frac { 0 }{ 0 } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule: Numerator derivative: \( \frac{d}{dx}(x - \sin x) = 1 - \cos x \) Denominator derivative: \( \frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x \)
\( \implies \lim _{x \rightarrow 0} \frac { 1 - \cos x }{ \sin x + x \cos x } \) Substitute \( x = 0 \): \( \frac { 1 - \cos(0) }{ \sin(0) + 0 \cdot \cos(0) } = \frac { 1 - 1 }{ 0 + 0 } = \frac { 0 }{ 0 } \) This is still an indeterminate form, so apply L' Hôpital's Rule again.
\( \implies \) Apply L' Hôpital's Rule again: Numerator derivative: \( \frac{d}{dx}(1 - \cos x) = \sin x \) Denominator derivative: \( \frac{d}{dx}(\sin x + x \cos x) = \cos x + (1 \cdot \cos x + x \cdot (-\sin x)) = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x \)
\( \implies \lim _{x \rightarrow 0} \frac { \sin x }{ 2 \cos x - x \sin x } \) Substitute \( x = 0 \): \( \frac { \sin(0) }{ 2 \cos(0) - 0 \cdot \sin(0) } = \frac { 0 }{ 2(1) - 0 } = \frac { 0 }{ 2 } = 0 \) The limit is 0. Combining fractions is a key step when dealing with \( \infty - \infty \) indeterminate forms.
In simple words: When we substitute \( x=0 \) into the problem, we get `infinity - infinity`, which is unclear. So, we first put the two fractions together into one. After that, we keep getting `0/0`, so we use L' Hôpital's Rule twice. This means we take the derivative of the top and bottom of the fraction, and then do it again. Finally, we get a clear answer of 0.

🎯 Exam Tip: Always remember to combine terms when dealing with \( \infty - \infty \) indeterminate forms to convert them into a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form before applying L' Hôpital's Rule.

Question 7. \( \lim _{x \rightarrow 1} \left( \frac { 2 }{ x^2-1 } – \frac { x }{ x-1 } \right) \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow 1} \left( \frac { 2 }{ x^2-1 } – \frac { x }{ x-1 } \right) \) Substitute \( x = 1 \): \( \frac { 2 }{ 1^2-1 } – \frac { 1 }{ 1-1 } = \frac { 2 }{ 0 } – \frac { 1 }{ 0 } = \infty - \infty \) This is an indeterminate form. We need to combine the fractions to apply L' Hôpital's Rule.
\( \implies \) Combine the fractions using a common denominator \( (x^2-1) \): We know that \( x^2-1 = (x-1)(x+1) \). So, \( \frac { x }{ x-1 } = \frac { x(x+1) }{ (x-1)(x+1) } = \frac { x(x+1) }{ x^2-1 } \)
\( \implies \frac { 2 }{ x^2-1 } – \frac { x(x+1) }{ x^2-1 } = \frac { 2 - x(x+1) }{ x^2-1 } \)
\( \implies = \frac { 2 - x^2 - x }{ x^2-1 } \)
\( \implies \) Now, apply the limit to the combined expression: \( \lim _{x \rightarrow 1} \frac { 2 - x^2 - x }{ x^2-1 } \) Substitute \( x = 1 \): \( \frac { 2 - 1^2 - 1 }{ 1^2-1 } = \frac { 2 - 1 - 1 }{ 1 - 1 } = \frac { 0 }{ 0 } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule: Numerator derivative: \( \frac{d}{dx}(2 - x^2 - x) = -2x - 1 \) Denominator derivative: \( \frac{d}{dx}(x^2-1) = 2x \)
\( \implies \lim _{x \rightarrow 1} \frac { -2x - 1 }{ 2x } \) Substitute \( x = 1 \): \( \frac { -2(1) - 1 }{ 2(1) } = \frac { -2 - 1 }{ 2 } = \frac { -3 }{ 2 } \) The limit is \( -\frac{3}{2} \). Always simplify the algebraic expressions carefully before applying L'Hopital's Rule.
In simple words: When we put \( x=1 \) into this problem, we get `infinity - infinity`, which is not a clear answer. To fix this, we combine the two fractions into one by finding a common bottom part. After combining, we get `0/0`, so we use L' Hôpital's Rule. We take the derivative of the top and bottom of the fraction, and then substitute \( x=1 \) again. The final answer is \( -\frac{3}{2} \).

🎯 Exam Tip: Factorize denominators like \( x^2-1 \) to find a common denominator more easily when combining fractions. This can simplify the subsequent differentiation steps for L' Hôpital's Rule.

Question 8. \( \lim _{x \rightarrow 0^ +} x^x \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow 0^ +} x^x \) Substitute \( x = 0 \): \( 0^0 \) This is an indeterminate form. We use logarithms to solve limits of the form \( f(x)^{g(x)} \).
\( \implies \) Let \( y = x^x \).
\( \implies \) Take the natural logarithm of both sides: \( \log y = \log (x^x) \) \( \log y = x \log x \)
\( \implies \) Now, evaluate the limit of \( \log y \): \( \lim _{x \rightarrow 0^ +} \log y = \lim _{x \rightarrow 0^ +} (x \log x) \) Substitute \( x = 0 \): \( 0 \cdot \log(0^+) = 0 \cdot (-\infty) \) This is an indeterminate form \( 0 \cdot \infty \). We need to rewrite it as a fraction.
\( \implies \) Rewrite as a fraction: \( x \log x = \frac { \log x }{ \frac{1}{x} } \)
\( \implies \) Now, apply the limit: \( \lim _{x \rightarrow 0^ +} \frac { \log x }{ \frac{1}{x} } \) Substitute \( x = 0 \): \( \frac { \log(0^+) }{ \frac{1}{0^+} } = \frac { -\infty }{ \infty } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule: Numerator derivative: \( \frac{d}{dx}(\log x) = \frac{1}{x} \) Denominator derivative: \( \frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2} \)
\( \implies \lim _{x \rightarrow 0^ +} \frac { \frac{1}{x} }{ -\frac{1}{x^2} } \)
\( \implies = \lim _{x \rightarrow 0^ +} \left( \frac{1}{x} \cdot (-x^2) \right) \)
\( \implies = \lim _{x \rightarrow 0^ +} (-x) \) Substitute \( x = 0 \): \( -0 = 0 \) So, \( \lim _{x \rightarrow 0^ +} \log y = 0 \).
\( \implies \) To find \( \lim _{x \rightarrow 0^ +} y \), we exponentiate the result: \( \lim _{x \rightarrow 0^ +} y = e^0 = 1 \) The limit is 1. Limits of the form \( f(x)^{g(x)} \) often require the use of logarithms and L'Hôpital's rule.
In simple words: We want to find the value of \( x^x \) as \( x \) gets very close to 0 from the positive side. When we try putting 0, we get `0 to the power of 0`, which is unclear. So, we use logarithms. We take the natural log of both sides, which changes the problem to `x log x`. After rearranging and using L' Hôpital's Rule, we find that `log y` goes to 0. Since `log y` is 0, `y` itself must be `e` to the power of 0, which is 1.

🎯 Exam Tip: For limits involving expressions of the form \( f(x)^{g(x)} \) that result in \( 0^0, \infty^0, \) or \( 1^\infty \) indeterminate forms, always use the method of taking the natural logarithm to convert them into \( 0 \cdot \infty \) or \( \infty - \infty \) forms.

Question 9. \( \lim _{x \rightarrow \infty} \left(1 + \frac { 1 }{ x }\right)^x \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow \infty} \left(1 + \frac { 1 }{ x }\right)^x \) Substitute \( x = \infty \): \( \left(1 + \frac { 1 }{ \infty }\right)^\infty = (1+0)^\infty = 1^\infty \) This is an indeterminate form. We use logarithms to solve limits of the form \( f(x)^{g(x)} \).
\( \implies \) Let \( y = \left(1 + \frac { 1 }{ x }\right)^x \).
\( \implies \) Take the natural logarithm of both sides: \( \log y = \log \left(1 + \frac { 1 }{ x }\right)^x \) \( \log y = x \log \left(1 + \frac { 1 }{ x }\right) \)
\( \implies \) Now, evaluate the limit of \( \log y \): \( \lim _{x \rightarrow \infty} \log y = \lim _{x \rightarrow \infty} \left(x \log \left(1 + \frac { 1 }{ x }\right)\right) \) Substitute \( x = \infty \): \( \infty \cdot \log \left(1 + \frac { 1 }{ \infty }\right) = \infty \cdot \log(1) = \infty \cdot 0 \) This is an indeterminate form \( \infty \cdot 0 \). We need to rewrite it as a fraction.
\( \implies \) Rewrite as a fraction: \( x \log \left(1 + \frac { 1 }{ x }\right) = \frac { \log \left(1 + \frac { 1 }{ x }\right) }{ \frac{1}{x} } \)
\( \implies \) Now, apply the limit: \( \lim _{x \rightarrow \infty} \frac { \log \left(1 + \frac { 1 }{ x }\right) }{ \frac{1}{x} } \) Substitute \( x = \infty \): \( \frac { \log \left(1 + 0\right) }{ 0 } = \frac { \log(1) }{ 0 } = \frac { 0 }{ 0 } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule: Let \( u = 1 + \frac{1}{x} \). Then \( \frac{du}{dx} = -\frac{1}{x^2} \). Numerator derivative: \( \frac{d}{dx}\left(\log \left(1 + \frac { 1 }{ x }\right)\right) = \frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) \) Denominator derivative: \( \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \)
\( \implies \lim _{x \rightarrow \infty} \frac { \frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) }{ -\frac{1}{x^2} } \)
\( \implies = \lim _{x \rightarrow \infty} \frac { 1 }{ 1 + \frac{1}{x} } \) Substitute \( x = \infty \): \( \frac { 1 }{ 1 + \frac{1}{\infty} } = \frac { 1 }{ 1 + 0 } = 1 \) So, \( \lim _{x \rightarrow \infty} \log y = 1 \).
\( \implies \) To find \( \lim _{x \rightarrow \infty} y \), we exponentiate the result: \( \lim _{x \rightarrow \infty} y = e^1 = e \) The limit is \( e \). This is a very important standard limit definition of the mathematical constant \( e \).
In simple words: We want to find the value of this special expression as \( x \) gets very, very large. When we directly substitute infinity, we get `1 to the power of infinity`, which is unclear. So, we use logarithms. We take the natural log of both sides, which helps us change the problem into a fraction. After using L' Hôpital's Rule, we find that `log y` goes to 1. This means `y` itself must be `e` to the power of 1, which is `e`.

🎯 Exam Tip: Recognize that \( \lim_{x \rightarrow \infty} \left(1 + \frac{a}{x}\right)^{bx} = e^{ab} \) is a fundamental limit. For this problem, \( a=1 \) and \( b=1 \), so the result is \( e^{1 \cdot 1} = e \).

Question 10. \( \lim _{x \rightarrow \frac{\pi}{2}} (\sin x)^{\tan x} \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow \frac{\pi}{2}} (\sin x)^{\tan x} \) Substitute \( x = \frac{\pi}{2} \): \( (\sin(\frac{\pi}{2}))^{\tan(\frac{\pi}{2})} = (1)^\infty \) This is an indeterminate form. We use logarithms to solve limits of the form \( f(x)^{g(x)} \).
\( \implies \) Let \( y = (\sin x)^{\tan x} \).
\( \implies \) Take the natural logarithm of both sides: \( \log y = \log ((\sin x)^{\tan x}) \) \( \log y = \tan x \log (\sin x) \)
\( \implies \) Now, evaluate the limit of \( \log y \): \( \lim _{x \rightarrow \frac{\pi}{2}} \log y = \lim _{x \rightarrow \frac{\pi}{2}} (\tan x \log (\sin x)) \) Substitute \( x = \frac{\pi}{2} \): \( \tan(\frac{\pi}{2}) \cdot \log(\sin(\frac{\pi}{2})) = \infty \cdot \log(1) = \infty \cdot 0 \) This is an indeterminate form \( \infty \cdot 0 \). We need to rewrite it as a fraction.
\( \implies \) Rewrite as a fraction: \( \tan x \log (\sin x) = \frac { \log (\sin x) }{ \frac{1}{\tan x} } = \frac { \log (\sin x) }{ \cot x } \)
\( \implies \) Now, apply the limit: \( \lim _{x \rightarrow \frac{\pi}{2}} \frac { \log (\sin x) }{ \cot x } \) Substitute \( x = \frac{\pi}{2} \): \( \frac { \log (\sin(\frac{\pi}{2})) }{ \cot(\frac{\pi}{2}) } = \frac { \log(1) }{ 0 } = \frac { 0 }{ 0 } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule: Numerator derivative: \( \frac{d}{dx}(\log (\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x \) Denominator derivative: \( \frac{d}{dx}(\cot x) = -\csc^2 x \)
\( \implies \lim _{x \rightarrow \frac{\pi}{2}} \frac { \cot x }{ -\csc^2 x } \) Substitute \( x = \frac{\pi}{2} \): \( \frac { \cot(\frac{\pi}{2}) }{ -\csc^2(\frac{\pi}{2}) } = \frac { 0 }{ -(1)^2 } = \frac { 0 }{ -1 } = 0 \) So, \( \lim _{x \rightarrow \frac{\pi}{2}} \log y = 0 \).
\( \implies \) To find \( \lim _{x \rightarrow \frac{\pi}{2}} y \), we exponentiate the result: \( \lim _{x \rightarrow \frac{\pi}{2}} y = e^0 = 1 \) The limit is 1. Converting \( \tan x \) to \( 1/\cot x \) helps to apply L'Hôpital's Rule effectively.
In simple words: We want to find the value of this problem as \( x \) gets very close to \( \frac{\pi}{2} \). When we put \( \frac{\pi}{2} \), we get `1 to the power of infinity`, which is unclear. So, we use logarithms. After taking the log, we rearrange the terms into a fraction and use L' Hôpital's Rule. We take the derivative of the top and bottom parts. The final result for `log y` is 0. This means the original expression `y` will be `e` to the power of 0, which is 1.

🎯 Exam Tip: When evaluating limits of \( f(x)^{g(x)} \), remember to convert \( \tan x \) to \( 1/\cot x \) or vice versa, depending on which form creates a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) indeterminate form more easily for L' Hôpital's Rule.

Question 11. \( \lim _{x \rightarrow 0^ +} (\cos x)^{\frac{1}{x^2}} \)
Answer:Let's evaluate the limit: Given limit: \( \lim _{x \rightarrow 0^ +} (\cos x)^{\frac{1}{x^2}} \) Substitute \( x = 0 \): \( (\cos(0))^{\frac{1}{0^2}} = (1)^\infty \) This is an indeterminate form. We use logarithms to solve limits of the form \( f(x)^{g(x)} \).
\( \implies \) Let \( y = (\cos x)^{\frac{1}{x^2}} \).
\( \implies \) Take the natural logarithm of both sides: \( \log y = \log ((\cos x)^{\frac{1}{x^2}}) \) \( \log y = \frac{1}{x^2} \log (\cos x) \)
\( \implies \) Now, evaluate the limit of \( \log y \): \( \lim _{x \rightarrow 0^ +} \log y = \lim _{x \rightarrow 0^ +} \left( \frac { \log (\cos x) }{ x^2 } \right) \) Substitute \( x = 0 \): \( \frac { \log (\cos(0)) }{ 0^2 } = \frac { \log(1) }{ 0 } = \frac { 0 }{ 0 } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule: Numerator derivative: \( \frac{d}{dx}(\log (\cos x)) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \) Denominator derivative: \( \frac{d}{dx}(x^2) = 2x \)
\( \implies \lim _{x \rightarrow 0^ +} \frac { -\tan x }{ 2x } \) Substitute \( x = 0 \): \( \frac { -\tan(0) }{ 2(0) } = \frac { 0 }{ 0 } \) This is still an indeterminate form, so apply L' Hôpital's Rule again.
\( \implies \) Apply L' Hôpital's Rule again: Numerator derivative: \( \frac{d}{dx}(-\tan x) = -\sec^2 x \) Denominator derivative: \( \frac{d}{dx}(2x) = 2 \)
\( \implies \lim _{x \rightarrow 0^ +} \frac { -\sec^2 x }{ 2 } \) Substitute \( x = 0 \): \( \frac { -\sec^2(0) }{ 2 } = \frac { -(1)^2 }{ 2 } = \frac { -1 }{ 2 } \) So, \( \lim _{x \rightarrow 0^ +} \log y = -\frac{1}{2} \).
\( \implies \) To find \( \lim _{x \rightarrow 0^ +} y \), we exponentiate the result: \( \lim _{x \rightarrow 0^ +} y = e^{-\frac{1}{2}} = \frac { 1 }{ e^{\frac{1}{2}} } = \frac { 1 }{ \sqrt{e} } \) The limit is \( \frac{1}{\sqrt{e}} \). The repeated application of L'Hôpital's Rule is essential for complex indeterminate forms.
In simple words: We are finding the value of this expression as \( x \) gets very close to 0 from the positive side. When we try to plug in 0, we get `1 to the power of infinity`, which is unclear. So, we use logarithms. We take the natural log of both sides and then arrange it as a fraction. Since we keep getting `0/0`, we apply L' Hôpital's Rule twice. After the second time, we get \( -\frac{1}{2} \) for `log y`. This means the original expression `y` will be `e` to the power of \( -\frac{1}{2} \), which is \( \frac{1}{\sqrt{e}} \).

🎯 Exam Tip: Be careful with the derivatives of trigonometric functions and powers. Mistakes in differentiation can lead to incorrect indeterminate forms and final answers. Double-check each derivative step.

Question 12. If an initial amount \( A_0 \) of money is invested at an interest rate \( r \) compounded \( n \) times a year, the value of the investment after \( t \) years is \( A = A_0\left(1 + \frac { r }{ n }\right)^{nt} \). If the interest is compounded continuously, (that is as \( n \rightarrow \infty \)), show that the amount after \( t \) years is \( A = A_0e^{rt} \).
Answer:We are given the formula for compound interest: \( A = A_0\left(1 + \frac { r }{ n }\right)^{nt} \) We need to find the amount \( A \) when interest is compounded continuously, which means taking the limit as \( n \rightarrow \infty \).
\( \implies A = \lim _{n \rightarrow \infty} A_0\left(1 + \frac { r }{ n }\right)^{nt} \) Since \( A_0 \) is a constant, we can take it out of the limit: \( A = A_0 \lim _{n \rightarrow \infty} \left(1 + \frac { r }{ n }\right)^{nt} \) Let \( y = \lim _{n \rightarrow \infty} \left(1 + \frac { r }{ n }\right)^{nt} \). Substitute \( n = \infty \): \( \left(1 + \frac { r }{ \infty }\right)^{\infty \cdot t} = (1+0)^\infty = 1^\infty \) This is an indeterminate form. We use logarithms to solve limits of the form \( f(x)^{g(x)} \).
\( \implies \) Let \( y = \left(1 + \frac { r }{ n }\right)^{nt} \).
\( \implies \) Take the natural logarithm of both sides: \( \log y = \log \left(\left(1 + \frac { r }{ n }\right)^{nt}\right) \) \( \log y = nt \log \left(1 + \frac { r }{ n }\right) \)
\( \implies \) Now, evaluate the limit of \( \log y \): \( \lim _{n \rightarrow \infty} \log y = \lim _{n \rightarrow \infty} \left(nt \log \left(1 + \frac { r }{ n }\right)\right) \) Substitute \( n = \infty \): \( \infty \cdot t \cdot \log \left(1 + \frac { r }{ \infty }\right) = \infty \cdot t \cdot \log(1) = \infty \cdot 0 \) This is an indeterminate form \( \infty \cdot 0 \). We need to rewrite it as a fraction.
\( \implies \) Rewrite as a fraction: \( nt \log \left(1 + \frac { r }{ n }\right) = \frac { t \log \left(1 + \frac { r }{ n }\right) }{ \frac{1}{n} } \)
\( \implies \) Now, apply the limit: \( \lim _{n \rightarrow \infty} \frac { t \log \left(1 + \frac { r }{ n }\right) }{ \frac{1}{n} } \) Substitute \( n = \infty \): \( \frac { t \log \left(1 + 0\right) }{ 0 } = \frac { t \cdot 0 }{ 0 } = \frac { 0 }{ 0 } \) This is an indeterminate form, so we can use L' Hôpital's Rule.
\( \implies \) Apply L' Hôpital's Rule (differentiate with respect to \( n \)): Let \( u = 1 + \frac{r}{n} \). Then \( \frac{du}{dn} = -\frac{r}{n^2} \). Numerator derivative: \( \frac{d}{dn}\left(t \log \left(1 + \frac { r }{ n }\right)\right) = t \cdot \frac{1}{1 + \frac{r}{n}} \cdot \left(-\frac{r}{n^2}\right) \) Denominator derivative: \( \frac{d}{dn}\left(\frac{1}{n}\right) = -\frac{1}{n^2} \)
\( \implies \lim _{n \rightarrow \infty} \frac { t \cdot \frac{1}{1 + \frac{r}{n}} \cdot \left(-\frac{r}{n^2}\right) }{ -\frac{1}{n^2} } \)
\( \implies = \lim _{n \rightarrow \infty} \frac { t \cdot r }{ 1 + \frac{r}{n} } \) Substitute \( n = \infty \): \( \frac { t \cdot r }{ 1 + \frac{r}{\infty} } = \frac { tr }{ 1 + 0 } = tr \) So, \( \lim _{n \rightarrow \infty} \log y = tr \).
\( \implies \) To find \( \lim _{n \rightarrow \infty} y \), we exponentiate the result: \( \lim _{n \rightarrow \infty} y = e^{tr} \) Substituting this back into the expression for \( A \): \( A = A_0 e^{rt} \) Hence Proved. This formula is vital in finance and continuous growth models.
In simple words: We start with the formula for how money grows with compound interest. We want to find out what happens when the interest is calculated all the time, not just a few times a year. This means we make `n` (the number of times compounded) go to infinity. We use a math trick with logarithms to solve this. After some steps, including using L' Hôpital's Rule, we find that the part that changes `y` becomes `e` to the power of `rt`. So, the final formula shows that the amount `A` equals the starting amount `A_0` multiplied by `e` to the power of `rt`. This is the formula for continuous compounding.

🎯 Exam Tip: This derivation is a classic result. Be familiar with the steps to transform the compound interest formula into the continuous compounding formula using limits and logarithms. Pay close attention to differentiating with respect to \( n \).