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

Practicing ML Aggarwal Class 12 Solutions ISC Chapter 6 Indeterminate Forms Chapter Test is the ultimate need for students who intend to score good marks in examinations.

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

Question 1.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 1} \frac{1-x^2}{\sin \pi x}\)
(ii) \(\ {Lt}_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{x-\cos \left(\sin ^{-1} x\right)}{1-\tan \left(\sin ^{-1} x\right)}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 1} \frac{1-x^2}{\sin \pi x}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{-2 x}{\pi \cos \pi x}\)
= \(\frac{-2}{\pi \cos \pi}\)
= \(\frac{-2}{\pi(-1)}=\frac{2}{\pi}\).

(ii) \(\ {Lt}_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{x-\cos \left(\sin ^{-1} x\right)}{1-\tan \left(\sin ^{-1} x\right)}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
put sin^-1 x = t
⇒ x = sin t
as x → \(\frac{1}{\sqrt{2}}\)
⇒ t → \(\frac{\pi}{4}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{t \rightarrow \frac{\pi}{4}} \frac{\sin t-\cos t}{1-\tan t}\)
= \(\ {Lt}_{t \rightarrow \frac{\pi}{4}} \frac{\cos t+\sin t}{-\sec ^2 t}\)
= \(\frac{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}{-\sec ^2 \frac{\pi}{4}}\)
= \(\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{-(\sqrt{2})^2}\)
= \(\frac{-\sqrt{2}}{2}=-\frac{1}{\sqrt{2}}\).

Question 2.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^2+x \log (1-x)}\)
(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x^x-x}{1-x+\log x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^2+x \log (1-x)}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]

(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x^x-x}{1-x+\log x}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{x^x(1+\log x)-1}{-1+\frac{1}{x}}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 1} \frac{x^x \cdot \frac{1}{x}+(1+\log x)^2 x^x}{-\frac{1}{x^2}}\)
= \(\frac{1 \times 1+(1 \times 0)^2 \cdot 1}{-1}\)
= – 2.

Question 3.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-x^2-2}{\sin ^2 x-x^2}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-x^2-2}{\sin ^2 x-x^2}\)
[\(\frac{0}{0}\) form, using L’Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 x}{\sin 2 x-2 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-2}{2 \cos 2 x-2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}}{-4 \sin 2 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}}{-8 \cos 2 x}\)
= \(\frac{1+1}{-8}=-\frac{1}{4}\).

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}\)

Question 4.
Evaluate the following limits:
(i) \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) log (1 – x) cot \(\frac{\pi}{2}\) x
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec² x – \(\frac{1}{x^2}\))
Solution:
(i) \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) log (1 – x) cot \(\frac{\pi}{2}\) x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log (1-x)}{\tan \frac{\pi}{2} x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\frac{1}{1-x}(-1)}{\frac{\pi}{2} \sec ^2 \frac{\pi x}{2}}\)
= \(-\frac{2}{\pi} \ {Lt}_{x \rightarrow 1^{-}} \frac{\cos ^2 \frac{\pi x}{2}}{1-x}\)
= \(\frac{2}{\pi} \ {Lt}_{x \rightarrow 1^{-}} \frac{2 \cos \frac{\pi x}{2}\left(-\sin \frac{\pi x}{2}\right) \frac{\pi}{2}}{-1}\)
= – \(\frac{2}{\pi}\) × 2 × 0 × 1 × \(\frac{\pi}{2}\) = 0.

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec² x – \(\frac{1}{x^2}\))

Question 5.
Evaluate the following limits:
(i) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\frac{4}{x^2}-\frac{2}{1-\cos x}\right)\)
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (x – sin^-1 x) cosec³ x
Solution:
(i) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\frac{4}{x^2}-\frac{2}{1-\cos x}\right)\) (∞ – ∞ form)

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (x – sin^-1 x) cosec³ x (0 . ∞ form)

Question 6.
Evaluate the following limits:
(i) \(\underset{x \rightarrow \frac{\pi}{2}}{\ {Lt}} \frac{\ {cosec} 6 x}{\ {cosec} 2 x}\)
(ii) \(\underset{x \rightarrow-\infty}{\mathbf{L t}}\) x² e^x
Solution:
(i) \(\underset{x \rightarrow \frac{\pi}{2}}{\ {Lt}} \frac{\ {cosec} 6 x}{\ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{\sin 2 x}{\sin 6 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos 2 x}{6 \cos 6 x}\)
= \(\frac{1}{3} \frac{\cos \pi}{\cos 3 \pi}=\frac{1}{3}\)

(ii) \(\underset{x \rightarrow-\infty}{\mathbf{L t}}\) x² e^x
= \(\ {Lt}_{x \rightarrow-\infty} \frac{x^2}{e^{-x}}\left(\frac{\infty}{\infty} \text { form }\right)\)
using L’Hopital’s rule
= \(\ {Lt}_{x \rightarrow-\infty} \frac{2 x}{-e^{-x}} \quad\left(\frac{\infty}{\infty} \text { form }\right)\)
= \(\ {Lt}_{x \rightarrow-\infty} \frac{2}{+e^{-x}}=\frac{2}{\infty} \rightarrow 0\)

Question 7.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 1}(2-x)^{\tan \frac{\pi x}{2}}\)
(ii) \(\ {Lt}_{x \rightarrow 1} x^{\frac{1}{x-1}}\)
Solution:
(i) Let y = \(\ {Lt}_{x \rightarrow 1}(2-x)^{\tan \frac{\pi x}{2}}\)
⇒ log y = tan \(\frac{\pi x}{2}\) log (2 – x)
∴ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) log y = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) tan \(\frac{\pi x}{2}\) log (2 – x) [0 . ∞ form]
= \(\ {Lt}_{x \rightarrow 1} \frac{\log (2-x)}{\cot \frac{\pi x}{2}}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 1} \frac{2 \sin ^2 \frac{\pi x}{2}}{\pi(2-x)}\)
= \(\frac{2}{\pi} \frac{(1)^2}{2-1}=\frac{2}{\pi}\)
log (\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y) = \(\frac{2}{\pi}\)
\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y = e^2/π
∴ \((2-x)^{\tan \frac{\pi x}{2}}\) = e^2/π

(ii) Let y = x^{\(\frac{1}{x-1}\)}
⇒ log y = \(\frac{\log x}{x-1}\)
∴ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) log y = \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) \(\frac{\log x}{x-1}\)
(\(\frac{0}{0}\) form)
using L’Hopital’s rule
= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) \(\frac{\frac{1}{x}}{1}=\frac{1}{1}\) = 1
∴ log (\(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y) = 1
⇒ \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) y = e’ = e
Thus, \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) x^{\(\frac{1}{x-1}\)} = e.

Question 8.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\log (3+x)-\log (3-x)}{x} & , x \neq 0 \\
k & , x=0
\end{array}\right.\), find the value of k so that the function f may be continuous at x = 0?
Solution:
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\log (3+x)-\log (3-x)}{x}\)
(\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{3+x}-\frac{1}{3-x}(-1)}{1}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{3-x+3+x}{(3+x)(3-x)}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{6}{9-x^2}=\frac{6}{9}=\frac{2}{3}\)
Also f(0) = k
Since the function f be continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ \(\frac{2}{3}\) = k.