ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.2

Parents can use ISC Maths Class 12 Solutions Chapter 6 Indeterminate Forms Ex 6.2 to provide additional support to their children.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.2

Evaluate the following (1 to 14) limits:

Question 1.
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \sin x}{\cot x}\)
(ii) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\tan x}{\log \cos x}\)
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \sin x}{\cot x}\) (\(\frac{\infty}{\infty}\) form)
using L’Hopital’s rule
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{\cos x}{\sin x}}{-\ {cosec}^2 x}\)
= – \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) cos x sin x
= – 1 × 0 = 0

(ii) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\tan x}{\log \cos x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sec ^2 x}{-\tan x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}}-\frac{1}{\cos ^2 x} \times \frac{\cos x}{\sin x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}} \frac{-1}{\cos x \sin x}\) → – ∞

Question 2.
(i) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{\log \left(x-\frac{\pi}{2}\right)}{\tan x}\)
(ii) \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log (1-x)}{\cot \pi x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{\log \left(x-\frac{\pi}{2}\right)}{\tan x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{+}}{2}} \frac{1}{\left(x-\frac{\pi}{2}\right) \sec ^2 x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{\cos ^2 x}{x-\frac{\pi}{2}}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{-2 \cos x \sin x}{1}\)
= – 2 × 0 × 1 = 0

(ii) \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log (1-x)}{\cot \pi x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{-1}{(1-x) \pi\left(-\ {cosec}^2 \pi x\right)}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\sin ^2 \pi x}{1-x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{2 \pi \sin \pi x \cos \pi x}{-1}\)

Question 3.
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \cot x}{e^{\ {cosec}^2 x}}\)
(ii) \(\underset{x \rightarrow 0^{+}}{\mathbf{L t}}\) log_{tan x} tan 2x.
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \cot x}{e^{\ {cosec}^2 x}}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\cot x}\left(-\ {cosec}^2 x\right)}{-e^{\ {cosec}^2 x} 2 \ {cosec}^2 x \cot x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{1}{2 \cot ^2 x e^{\ {cosec}^2 x}}\)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{1}{2}\) tan² x e^{– cosec2 x}
= \(\frac{1}{2}\) × 0 × 0 = 0
[∵ e^{– cosec2 x} → e^{– ∞} = 0, ae, x → 0⁺]

(ii) \(\underset{x \rightarrow 0^{+}}{\mathbf{L t}}\) log_{tan x} tan 2x

Question 4.
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log x}{x}\)
(ii) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{5 x+2 \log x}{x+3 \log x}\)
Solution:
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log x}{x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{x}}{1}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{1}{x}\) = 0.

(ii) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{5 x+2 \log x}{x+3 \log x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \infty} \frac{5+\frac{2}{x}}{1+\frac{3}{x}}\)
= \(\frac{5+0}{1+0}\)
= 5

Question 5.
(i) \(\ {Lt}_{x \rightarrow \infty} \frac{x^4+x^2}{e^x+1}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\cot x}{\cot 2 x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow \infty} \frac{x^4+x^2}{e^x+1}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\cot x}{\cot 2 x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{\tan 2 x}{\tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \sec ^2 2 x}{\sec ^2 x}\)
= \(\frac{2 \times 1^2}{1^2}\) = 2.

Question 6.
(i) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) cosec πx log x
(ii) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) (1 – x) tan \(\frac{\pi}{2}\) x
Solution:
(i) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) cosec πx log x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 1} \frac{\log x}{\sin \pi x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{\frac{1}{x}}{\pi \cos \pi x}\)
= \(\frac{1}{1 \times \pi \times \cos \pi}=\frac{1}{-\pi}\)

(ii) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) (1 – x) tan \(\frac{\pi}{2}\) x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 1} \frac{1-x}{\cot \frac{\pi x}{2}}\) [\(\frac{0}{0}\) form]
= \(\ {Lt}_{x \rightarrow 1} \frac{-1}{-\frac{\pi}{2} \ {cosec}^2 \frac{\pi x}{2}}\)
= \(\ {Lt}_{x \rightarrow 1} \frac{2}{\pi} \sin ^2 \frac{\pi x}{2}\)
= \(\frac{2}{\pi} \sin ^2 \frac{\pi}{2}\)
= \(\frac{2}{\pi}(1)^2\)
= \(\frac{2}{\pi}\).

Question 7.
(i) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{x^2-1}-\frac{2}{x^4-1}\right)\)
(ii) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{\log x}-\frac{1}{x-1}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{x^2-1}-\frac{2}{x^4-1}\right)\) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow 1}\left[\frac{x^2+1-2}{x^4-1}\right]\)
= \(\ {Lt}_{x \rightarrow 1} \frac{x^2-1}{x^4-1}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{2 x}{4 x^3}\)
= \(\frac{2}{4}=\frac{1}{2}\)

(ii) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{\log x}-\frac{1}{x-1}\right)\)

Question 8.
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^2} \log (1+x)\right)\)
(ii) \(\ {Lt}_{x \rightarrow 2}\left(\frac{1}{\log (x-1)}-\frac{1}{x-2}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^2} \log (1+x)\right)\) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow 0}\left[\frac{x-\log (1+x)}{x^2}\right]\) (\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{1-\frac{1}{1+x}}{2 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{1+x-1}{2 x(1+x)}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{1}{(1+x)^2}\)
= \(\frac{1}{2(1+0)}=\frac{1}{2}\)

(ii) \(\ {Lt}_{x \rightarrow 2}\left(\frac{1}{\log (x-1)}-\frac{1}{x-2}\right)\) (∞ – ∞ form)

Question 9.
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)\)
(ii) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{2 x}-\frac{1}{x\left(e^{\pi x}+1\right)}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)\) (∞ – ∞ form)

 

(ii) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{2 x}-\frac{1}{x\left(e^{\pi x}+1\right)}\right)\) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow 0}\left[\frac{e^{\pi x}+1-2}{2 x\left(e^{\pi x}+1\right)}\right]\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x}-1}{2 x\left(e^{\pi x}+1\right)}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x}-1}{2 x} \cdot \frac{1}{1+1}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x}-1}{4 x}\) (\(\frac{0}{0}\) form)
using L’Hopital’s rule, we have
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x} \cdot \pi}{4}=\frac{\pi}{4}\)

Question 10.
(i) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\cot x-\frac{1}{x}\right)\)
(ii) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\cot ^2 x-\frac{1}{x^2}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\cot x-\frac{1}{x}\right)\) (∞ – ∞ form)

 

(ii) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\cot ^2 x-\frac{1}{x^2}\right)\)

 

Question 11.
(i) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (sec x – tan x)
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec x – cot x)
Solution:
(i) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (sec x – tan x) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}}\left[\frac{1-\sin x}{\cos x}\right]\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\sin x}\)
= \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) cot x = 0.

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec x – cot x)
= \(\ {Lt}_{x \rightarrow 0} \frac{1-\cos x}{\sin x}\) (\(\frac{0}{0}\) form)
using ‘L’ Hopitals’s rule
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin x}{\cos x}\)
= \(\frac{0}{1}\) = 0.

Question 12.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log x
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) sin x log x².
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\log x}{\frac{1}{x}}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (- x) = 0.

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) sin x log x²
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 2 sin x log x (0 . ∞ form)
[∵ log a^b = b log a]

 

Question 13.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log (tan x)
(ii) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (1 – sin x) tan x
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log (tan x) (0 . ∞ form)

(ii) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (1 – sin x) tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cot x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\ {cosec}^2 x}\)
= \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) cos x sin² x
= 0 × 1² = 0.

Question 14.
(i) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x tan (\(\frac{\pi}{2}\) – x)
(ii) \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (c – x) tan \(\frac{\pi x}{2 c}\)
Solution:
(i) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x tan (\(\frac{\pi}{2}\) – x)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x cot x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0} \frac{x}{\tan x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{1}{\sec ^2 x}\)
= \(\frac{1}{1^2}\) = 1.

(ii) \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (c – x) tan \(\frac{\pi x}{2 c}\) (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow c} \frac{c-x}{\cot \frac{\pi x}{2 c}}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow c} \frac{-1}{-\ {cosec}^2 \frac{\pi x}{2 c} \cdot \frac{\pi}{2 c}}\)
= \(\frac{2 c}{\pi} \ {Lt}_{x \rightarrow c} \sin ^2 \frac{\pi x}{2 c}\)
= \(\frac{2 c}{\pi} \times 1^2\)
= \(\frac{2 c}{\pi}\)

Question 15.
Show that \(\ {Lt}_{x \rightarrow \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\) = 1 and \(\ {Lt}_{x \rightarrow-\infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\) = – 1.
Solution:
\(\ {Lt}_{x \rightarrow \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \infty} \frac{e^x\left(1-e^{-2 x}\right)}{e^x\left(1+e^{-2 x}\right)}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{1-e^{-2 x}}{1+e^{-2 x}}\)
= \(\frac{1-0}{1+0}\) = 1
[as x → ∞, e^-2x → e^{– ∞} = \(\frac{1}{e^{\infty}}\) = 0]
and \(\ {Lt}_{x \rightarrow-\infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\)
= \(\ {Lt}_{x \rightarrow-\infty} \frac{e^{2 x}-1}{e^{2 x}+1}\)
= \(\frac{0-1}{0+1}\)
= 1.