ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.3

Students can track their progress and improvement through regular use of Class 12 ISC Maths Solutions Chapter 6 Indeterminate Forms Ex 6.3.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.3

Evaluate the following (1 to 6) limits:

Question 1.
(i) \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) (x – α)^{x – α}
(ii) \(\underset{x \rightarrow 1^{-}}{\mathbf{L t}}\) \(\left(1-x^2\right)^{\frac{1}{\log (1-x)}}\)
S0lution:
(i) Let F(x) = (x – α)^{x – α}
⇒ log F(x) = (x – α) log (x – α)
∴ \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) log (F(x)) = \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) (x – α) log (x – α)
= \(\ {Lt}_{x \rightarrow \alpha} \frac{\log (x-\alpha)}{\frac{1}{x-\alpha}}\)
= \(\ {Lt}_{x \rightarrow \alpha} \frac{\frac{1}{x-\alpha}}{-\frac{1}{(x-\alpha)^2}}\)
= \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) – (x – α) = 0
⇒ log (\(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) (x – α)^{x – α} = e⁰ = 1

(ii) Let F(x) = \(\underset{x \rightarrow 1^{-}}{\mathbf{L t}}\) \(\left(1-x^2\right)^{\frac{1}{\log (1-x)}}\)
⇒ log F(x) = \(\frac{\log \left(1-x^2\right)}{\log (1-x)}\)
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) log (F(x)) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log \left(1-x^2\right)}{\log (1-x)}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\frac{1}{1-x^2}(-2 x)}{\frac{1}{1-x}(-1)}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{2 x(1-x)}{1-x^2}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{2 x}{1+x}\)
∴ log (\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) F(x)) = 1
⇒ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) F(x) = e
⇒ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) \(\left(1-x^2\right)^{\frac{1}{\log (1-x)}}\) = e.

Question 2.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (tan x)^{sin 2x}
(ii) \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) (tan x)^{sin 2x}
Solution:
(i) Let F(x) = sin 2x log tan x
⇒ log F(x) = sin 2x log tan x
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log (F(x)) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) sin 2x log tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\log \tan x}{\ {cosec} 2 x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\tan x} \sec ^2 x}{-\cot 2 x \ {cosec} 2 x \cdot 2}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{2}{\frac{\sin 2 x}{-2 \cot 2 x \{cosec} {2 x}}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\ {cosec} 2 x}{-\cot 2 x \ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow 0}\) (- tan 2x) = 0
⇒ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e⁰ = 1
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (tan x)^{sin 2x} = 1

(ii) Let F(x) = (tan x)^{sin 2x}
⇒ log F(x) = sin 2x log tan x
∴ \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) F(x) = \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) sin 2x log tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}} \frac{\log \tan x}{\ {cosec} 2 x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\frac{\sec ^2 x}{\tan x}}{2 \cot 2 x \ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{-}} \frac{2 \ {cosec} 2 x}{-2 \cot 2 x \ {cosec} 2 x}\)
= – \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) tan 2x = 0
⇒ log (\(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) F(x) = e⁰ = 1
⇒ \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) (tan x)^{sin 2x} = 1.

Question 3.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \((\cot x)^{\frac{1}{\log x}}\)
(ii) \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) (1 + x)^1/x
Solution:
Let F(x) = \((\cot x)^{\frac{1}{\log x}}\)
⇒ log f(x) = \(\frac{\log \cot x}{\log x}\)
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\frac{\log \cot x}{\log x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\cot x}\left(-\ {cosec}^2 x\right)}{\frac{1}{x}}\) [using L ‘Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{-x \tan x}{\sin ^2 x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{-x}{\cos x \sin x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x}{\sin x} \cdot \underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{1}{\cos x}\)
∴ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x)) = – 1. 1 = – 1
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e^-1
= \(\frac{1}{e}\)
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \((\cot x)^{\frac{1}{\log x}}\) = \(\frac{1}{e}\).

(ii) Let F(x) = (1 + x)^1/x
⇒ log F(x) = \(\frac{1}{x}\) log (1 + x)
\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) log F(x) = \(\ {Lt}_{x \rightarrow \infty} \frac{\log (1+x)}{x}\) (\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{1+x}}{1}=\frac{1}{\infty}\) = 0
∴ log (\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) F(x) = e⁰ = 1
\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) (1 + x)^1/x = 1.

Question 4.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\left(\frac{1}{x}\right)^{\tan x}\)
(ii) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (cot x)^{sin 2x}
Solution:
(i) Let f(x) = \(\left(\frac{1}{x}\right)^{\tan x}\)
⇒ log F(x) = tan x log (\(\frac{1}{x}\))
= – tan x log x
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x) = – \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) tan x log x
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) – \(\frac{\log x}{\cot x} \quad\left(\frac{\infty}{\infty}\right)\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{-\frac{1}{x}}{-\ {cosec}^2 x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\sin ^2 x}{x}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{2 \sin x \cos x}{1}\)
= 2 × 0 × 1 = 0
∴ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x)) = 0
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e⁰ = 1
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\left(\frac{1}{x}\right)^{\tan x}\) = 1

(ii) Let F(x) = (cot x)^{sin 2x}
⇒ log F(x) = sin 2x log cot x
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) sin 2x log cot x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\log \cot x}{\ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\cot x}\left(-\ {cosec}^2 x\right)}{-2 \ {cosec} 2 x \cot 2 x}\)
= \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{2 \ {cosec} 2 x}{2 \ {cosec} 2 x \cot 2 x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) tan 2x = 0
∴ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e⁰ = 1
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (cot x)^{sin 2x} = 1.

Question 5.
(i) \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) (sec x)^{cot x}
(ii) \(\begin{gathered}
\text { Lt } \\
x \rightarrow \frac{\pi}{4}
\end{gathered}\) (tan x)^{tan 2x}
Solution:
(i) Let F(x) = (sec x)^{cot x}
⇒ log F(x) = \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) cot x log sec x
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\log (\sec x)}{\tan x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\frac{1}{\sec x} \sec x \tan x}{\sec ^2 x}\)
= \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) cos x sin x
= 0 × 1 = 0
⇒ log (\(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) F(x) = e⁰ = 1
⇒ \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) (sec x)^{cot x} = 1

(ii) Let F(x) = (tan x)^{tan 2x}
⇒ log F(x) = tan 2x log tan x
\(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) log F(x) = \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) tan 2x log tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{4}} \frac{\log \tan x}{\cot 2 x}\left(\frac{0}{0} \text { form }\right)\)
using L’Hopital’s rule, we have
= \(\ {Lt}_{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\tan x} \sec ^2 x}{-2 \ {cosec}^2 2 x}\)
= \(\frac{\frac{1}{1} \cdot(\sqrt{2})^2}{-2(1)^2}\) = – 1
⇒ log (\(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) F(x)) = – 1
⇒ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) F(x) = e^-1 = \(\frac{1}{e}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) (tan x)^{tan 2x} = \(\frac{1}{e}\).

Question 6.
(i) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (1 + x)^1/x
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (e^x + 4x)^1/x
Solution:
(i) Let F(x) = (1 + x)^1/x
⇒ log F(x) = \(\frac{\log (1+x)}{x}\)
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\log (1+x)}{x}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{i \rightarrow 0} \frac{1}{\frac{1+x}{1}}\)
= \(\frac{1}{1+0}\) = 1
⇒ log (\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x)) = 1
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x) = e⁰ = 1
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (1 + x)^1/x = 1.

(ii) Let F(x) = (e^x + 4x)^1/x
⇒ log F(x) = \(\frac{1}{x}\) log (e^x + 4x)
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\log \left(e^x+4 x\right)}{x}\) (\(\frac{0}{0}\) form)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\frac{1}{e^x+4 x}\left(e^x+4\right)}{1}\)
= \(\frac{e^0+4}{e^0+4 \times 0}\) = 5
⇒ log (\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x)) = 5
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x) = e⁵
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (e^x + 4x) = e⁵.