OP Malhotra Class 12 Maths Solutions Chapter 9 Indeterminate Forms of Limits Chapter Test

Question 2. \( \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sec x}{1+\tan x} \)
Answer: To find the limit, we first substitute \( x = \frac{\pi}{2} \). Since \( \sec \frac{\pi}{2} \) and \( \tan \frac{\pi}{2} \) are undefined, this is an indeterminate form \( \frac{\infty}{\infty} \). We can use L'Hopital's rule, which means we differentiate the numerator and the denominator separately.
\[ \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sec x}{1+\tan x} \]
\( \implies \) Differentiating the numerator: \( \frac{d}{dx}(\sec x) = \sec x \tan x \)
\( \implies \) Differentiating the denominator: \( \frac{d}{dx}(1+\tan x) = \sec^2 x \)
\( \implies \) Now, apply L'Hopital's Rule:
\( \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sec x \tan x}{\sec^2 x} \)
\( \implies \) Simplify the expression:
\( \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan x}{\sec x} \)
\( \implies \) Convert to sine and cosine: \( \frac{\sin x / \cos x}{1 / \cos x} = \sin x \)
\( \implies \lim _{x \rightarrow \frac{\pi}{2}} \sin x \)
\( \implies \) Substitute \( x = \frac{\pi}{2} \):
\( = \sin \frac{\pi}{2} \)
\( = 1 \) The value of the limit is 1.
In simple words: When you try to put \( x = \frac{\pi}{2} \) into the function, you get an uncertain answer like infinity divided by infinity. So, we use a special rule called L'Hopital's rule: we take the derivative of the top part and the derivative of the bottom part. After simplifying, we get just \( \sin x \), and putting \( x = \frac{\pi}{2} \) into that gives us 1.

🎯 Exam Tip: Always identify the indeterminate form (like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)) first to know if L'Hopital's Rule is applicable. Simplify the expression after differentiation to make the final substitution easier.

Question 3. \( \lim_{x \rightarrow \infty} \frac{x-2x^2}{3 x^2+5 x} \)
Answer: To find the limit as \( x \) approaches infinity, we first substitute infinity into the expression, which gives us an indeterminate form \( \frac{\infty}{\infty} \). To solve this, we can divide both the numerator and the denominator by the highest power of \( x \) present, which is \( x^2 \).
\[ \lim_{x \rightarrow \infty} \frac{x-2x^2}{3 x^2+5 x} \]
\( \implies \) Divide numerator and denominator by \( x^2 \):
\( \lim_{x \rightarrow \infty} \frac{\frac{x}{x^2}-\frac{2x^2}{x^2}}{\frac{3x^2}{x^2}+\frac{5x}{x^2}} \)
\( \implies \lim_{x \rightarrow \infty} \frac{\frac{1}{x}-2}{3+\frac{5}{x}} \)
\( \implies \) Now, substitute \( x = \infty \). As \( x \rightarrow \infty \), \( \frac{1}{x} \rightarrow 0 \) and \( \frac{5}{x} \rightarrow 0 \).
\( = \frac{0-2}{3+0} \)
\( = \frac{-2}{3} \) The value of the limit is \( -\frac{2}{3} \).
In simple words: When \( x \) becomes very, very big (goes to infinity), we need to find out what number the whole expression gets close to. Since both the top and bottom also get very big, we divide every term by the largest power of \( x \). After doing that, any part with \( x \) in the bottom will become zero, leaving us with just the numbers.

🎯 Exam Tip: For limits involving \( x \rightarrow \infty \) with rational functions, always divide by the highest power of \( x \) in the denominator to simplify the expression. Terms like \( \frac{c}{x^n} \) will tend to 0 as \( x \rightarrow \infty \).

Question 4. \( \lim _{x \rightarrow 0} \frac{\sin x-x}{x^2} \)
Answer: To find the limit, we first substitute \( x=0 \). This results in \( \frac{\sin 0 - 0}{0^2} = \frac{0}{0} \), which is an indeterminate form. We must use L'Hopital's rule, where we differentiate the numerator and the denominator separately until the indeterminate form is resolved.
\[ \lim _{x \rightarrow 0} \frac{\sin x-x}{x^2} \]
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(\sin x-x) = \cos x - 1 \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(x^2) = 2x \)
\( \implies \) Apply L'Hopital's Rule once:
\( \lim _{x \rightarrow 0} \frac{\cos x-1}{2x} \)
\( \implies \) Substitute \( x=0 \) again: \( \frac{\cos 0 - 1}{2 \times 0} = \frac{1-1}{0} = \frac{0}{0} \), still an indeterminate form.
\( \implies \) Apply L'Hopital's Rule a second time.
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(\cos x-1) = -\sin x \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(2x) = 2 \)
\( \implies \) Now, apply L'Hopital's Rule a second time:
\( \lim _{x \rightarrow 0} \frac{-\sin x}{2} \)
\( \implies \) Substitute \( x=0 \):
\( = \frac{-\sin 0}{2} \)
\( = \frac{0}{2} \)
\( = 0 \) The value of the limit is 0.
In simple words: When we put \( x=0 \) into the problem, we get zero divided by zero, which doesn't tell us the answer. So, we use L'Hopital's rule: we take the derivative of the top and bottom. We have to do this twice. After the second time, we put \( x=0 \) in again, and we get the final answer.

🎯 Exam Tip: Remember to apply L'Hopital's Rule multiple times if you still get an indeterminate form after the first differentiation. Also, ensure you differentiate the numerator and denominator independently, not as a quotient rule.

Question 5. \( \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2} \)
Answer: To find the limit, we first substitute \( \theta=0 \). This gives \( \frac{1-\cos 0}{0^2} = \frac{1-1}{0} = \frac{0}{0} \), which is an indeterminate form. We need to use L'Hopital's Rule, which means differentiating the numerator and the denominator separately.
\[ \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2} \]
\( \implies \) Differentiate numerator: \( \frac{d}{d\theta}(1-\cos \theta) = \sin \theta \)
\( \implies \) Differentiate denominator: \( \frac{d}{d\theta}(\theta^2) = 2\theta \)
\( \implies \) Apply L'Hopital's Rule:
\( \lim _{\theta \rightarrow 0} \frac{\sin \theta}{2\theta} \)
\( \implies \) We can rewrite this as:
\( \frac{1}{2} \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \)
\( \implies \) We know that \( \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \). This is a standard limit in trigonometry.
\( = \frac{1}{2} \times 1 \)
\( = \frac{1}{2} \) The value of the limit is \( \frac{1}{2} \).
In simple words: If you try to put zero into this problem, you get zero divided by zero. So, we use L'Hopital's rule, which means we differentiate the top and bottom. This gives us a form with \( \sin \theta \) over \( \theta \), and we know that this part becomes 1 when \( \theta \) is zero. So, the final answer is \( \frac{1}{2} \).

🎯 Exam Tip: Recognize the standard limit \( \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 \), as it often appears after applying L'Hopital's Rule in trigonometric limit problems. This helps simplify the solution quickly.

Question 6. \( \lim _{x \rightarrow 0} \frac{e^x-(1+x)}{x^2} \)
Answer: To find the limit, we first substitute \( x=0 \). This results in \( \frac{e^0-(1+0)}{0^2} = \frac{1-1}{0} = \frac{0}{0} \), which is an indeterminate form. We must use L'Hopital's rule by differentiating the numerator and the denominator separately.
\[ \lim _{x \rightarrow 0} \frac{e^x-(1+x)}{x^2} \]
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(e^x-1-x) = e^x - 1 \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(x^2) = 2x \)
\( \implies \) Apply L'Hopital's Rule once:
\( \lim _{x \rightarrow 0} \frac{e^x-1}{2x} \)
\( \implies \) Substitute \( x=0 \) again: \( \frac{e^0-1}{2 \times 0} = \frac{1-1}{0} = \frac{0}{0} \), which is still an indeterminate form.
\( \implies \) Apply L'Hopital's Rule a second time.
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(e^x-1) = e^x \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(2x) = 2 \)
\( \implies \) Now, apply L'Hopital's Rule a second time:
\( \lim _{x \rightarrow 0} \frac{e^x}{2} \)
\( \implies \) Substitute \( x=0 \):
\( = \frac{e^0}{2} \)
\( = \frac{1}{2} \) The value of the limit is \( \frac{1}{2} \).
In simple words: If you try to put zero into this equation, you get zero divided by zero. This means we need to use a rule called L'Hopital's rule. We have to do this twice, taking the derivative of the top and bottom parts each time. After the second time, we can finally put zero in and get the answer, which is \( \frac{1}{2} \).

🎯 Exam Tip: When using L'Hopital's Rule for exponential functions, remember that \( \frac{d}{dx}(e^x) = e^x \). Be careful with the signs and constants when differentiating the terms in the numerator and denominator.

Question 7. \( \lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2} \)
Answer: To find the limit, we first substitute \( x=1 \). This gives \( \frac{1+\log 1-1}{1-2(1)+1^2} = \frac{1+0-1}{1-2+1} = \frac{0}{0} \), an indeterminate form. We need to use L'Hopital's Rule by differentiating the numerator and the denominator separately.
\[ \lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2} \]
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(1+\log x-x) = \frac{1}{x} - 1 \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(1-2x+x^2) = -2+2x \)
\( \implies \) Apply L'Hopital's Rule:
\( \lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2x} \)
\( \implies \) Simplify the numerator: \( \frac{1-x}{x} \)
\( \implies \) Factor the denominator: \( 2(x-1) \)
\( \implies \lim _{x \rightarrow 1} \frac{\frac{1-x}{x}}{2(x-1)} \)
\( \implies \lim _{x \rightarrow 1} \frac{-(x-1)}{x \cdot 2(x-1)} \)
\( \implies \) Cancel out the common factor \( (x-1) \):
\( \lim _{x \rightarrow 1} \frac{-1}{2x} \)
\( \implies \) Substitute \( x=1 \):
\( = \frac{-1}{2(1)} \)
\( = -\frac{1}{2} \) The value of the limit is \( -\frac{1}{2} \).
In simple words: When we put \( x=1 \) into this problem, we get zero divided by zero. So we use L'Hopital's rule, which means we differentiate the top and bottom parts. After differentiating, we simplify the expression and cancel out common parts. Then, we put \( x=1 \) back in to get the final answer.

🎯 Exam Tip: After applying L'Hopital's Rule, always try to simplify the expression by factoring or algebraic manipulation before substituting the limit value. This often helps to cancel out terms causing the indeterminate form.

Question 8. \( \lim _{x \rightarrow 0}(\cot x)^{\frac{1}{\log x}} \)
Answer: This limit is of the form \( \infty^0 \) as \( x \rightarrow 0 \) (since \( \cot 0 = \infty \) and \( \log 0 = -\infty \), so \( \frac{1}{\log 0} = 0 \)). For limits of the form \( f(x)^{g(x)} \), we use logarithms. Let \( y = \lim _{x \rightarrow 0}(\cot x)^{\frac{1}{\log x}} \).
\( \implies \log y = \lim _{x \rightarrow 0} \log \left( (\cot x)^{\frac{1}{\log x}} \right) \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{1}{\log x} \log(\cot x) \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\log(\cot x)}{\log x} \)
\( \implies \) As \( x \rightarrow 0 \), this is of the form \( \frac{\infty}{\infty} \) (since \( \cot x \rightarrow \infty \Rightarrow \log(\cot x) \rightarrow \infty \), and \( \log x \rightarrow -\infty \)). So, we can apply L'Hopital's Rule.
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(\log(\cot x)) = \frac{1}{\cot x} (-\csc^2 x) = \tan x (-\csc^2 x) = -\frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = -\frac{1}{\sin x \cos x} \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(\log x) = \frac{1}{x} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-\frac{1}{\sin x \cos x}}{\frac{1}{x}} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-x}{\sin x \cos x} \)
\( \implies \) We know that \( \sin(2x) = 2 \sin x \cos x \). So, \( \sin x \cos x = \frac{1}{2} \sin(2x) \).
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-x}{\frac{1}{2} \sin(2x)} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-2x}{\sin(2x)} \)
\( \implies \) This is of the form \( \frac{0}{0} \). We can use the standard limit \( \lim_{u \rightarrow 0} \frac{u}{\sin u} = 1 \). Here, let \( u = 2x \).
\( \implies \log y = -1 \times \lim _{x \rightarrow 0} \frac{2x}{\sin(2x)} \)
\( \implies \log y = -1 \times 1 \)
\( \implies \log y = -1 \)
\( \implies y = e^{-1} \)
\( \implies y = \frac{1}{e} \) The value of the limit is \( \frac{1}{e} \).
In simple words: When we have a function raised to another function, and it gives an uncertain form like infinity to the power of zero, we use logarithms to change it. We take the logarithm of both sides, which brings the power down. Then we apply L'Hopital's rule by differentiating the top and bottom. After simplifying and using a known trigonometric limit, we find the logarithm of our answer, and then we raise \( e \) to that power to get the final answer.

🎯 Exam Tip: For limits of the form \( f(x)^{g(x)} \) that result in indeterminate forms like \( 0^0, \infty^0, 1^\infty \), always start by taking the natural logarithm of both sides and converting the expression to a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form to apply L'Hopital's Rule.

Question 9. \( \lim _{x \rightarrow 0}(1-2 x)^{1 / x} \)
Answer: This limit is of the form \( 1^\infty \) as \( x \rightarrow 0 \) (since \( (1-2 \times 0) = 1 \) and \( \frac{1}{0} = \infty \)). For limits of the form \( f(x)^{g(x)} \), we use logarithms. Let \( y = \lim _{x \rightarrow 0}(1-2 x)^{1 / x} \).
\( \implies \log y = \lim _{x \rightarrow 0} \log \left( (1-2x)^{1/x} \right) \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{1}{x} \log(1-2x) \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\log(1-2x)}{x} \)
\( \implies \) As \( x \rightarrow 0 \), this is of the form \( \frac{\log(1)}{0} = \frac{0}{0} \). So, we can apply L'Hopital's Rule.
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(\log(1-2x)) = \frac{1}{1-2x}(-2) = \frac{-2}{1-2x} \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(x) = 1 \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{\frac{-2}{1-2x}}{1} \)
\( \implies \log y = \lim _{x \rightarrow 0} \frac{-2}{1-2x} \)
\( \implies \) Substitute \( x=0 \):
\( = \frac{-2}{1-2(0)} \)
\( = \frac{-2}{1} \)
\( = -2 \)
\( \implies \log y = -2 \)
\( \implies y = e^{-2} \) The value of the limit is \( e^{-2} \).
In simple words: When we see a number that gets close to 1 raised to a power that gets very big, we have an uncertain answer. To solve it, we use logarithms. We put \( \log \) on both sides, which helps move the power down. Then we differentiate the top and bottom parts. Finally, we put the limit value back in, and we get the logarithm of our answer. We then take \( e \) to that power to find the true limit.

🎯 Exam Tip: Be careful with the chain rule when differentiating \( \log(1-2x) \). Remember to multiply by the derivative of the inner function, which is \( -2 \).

Question 10. \( \lim _{x \rightarrow 1} \frac{\tan (x^2-1)}{x-1} \)
Answer: To find the limit, we first substitute \( x=1 \). This gives \( \frac{\tan (1^2-1)}{1-1} = \frac{\tan 0}{0} = \frac{0}{0} \), which is an indeterminate form. We need to use L'Hopital's Rule, differentiating the numerator and the denominator separately.
\[ \lim _{x \rightarrow 1} \frac{\tan (x^2-1)}{x-1} \]
\( \implies \) Differentiate numerator: \( \frac{d}{dx}(\tan (x^2-1)) = \sec^2 (x^2-1) \cdot (2x) \)
\( \implies \) Differentiate denominator: \( \frac{d}{dx}(x-1) = 1 \)
\( \implies \) Apply L'Hopital's Rule:
\( \lim _{x \rightarrow 1} \frac{\sec^2 (x^2-1) \cdot 2x}{1} \)
\( \implies \) Substitute \( x=1 \):
\( = \sec^2 (1^2-1) \cdot 2(1) \)
\( = \sec^2 (0) \cdot 2 \)
\( \implies \) We know that \( \sec 0 = 1 \).
\( = (1)^2 \cdot 2 \)
\( = 1 \cdot 2 \)
\( = 2 \) The value of the limit is 2.
In simple words: When we try to put \( x=1 \) into this problem, we get zero divided by zero. So we use L'Hopital's rule, which means we differentiate the top and bottom parts. After differentiating, we put \( x=1 \) back into the new expression. We use the fact that \( \sec 0 \) is 1 to find the final answer.

🎯 Exam Tip: Remember the derivative of \( \tan u \) is \( \sec^2 u \cdot \frac{du}{dx} \) by the chain rule. Also, be familiar with basic trigonometric values like \( \sec 0 = 1 \).