The availability of ISC Mathematics Class 12 Solutions Chapter 6 Indeterminate Forms Ex 6.1 encourages students to tackle difficult exercises.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.1

Evaluate the following (1 to 13) limits:

Question 1.
(i) $$\ {Lt}_{x \rightarrow 3} \frac{x^4-81}{x-3}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-1}{x}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 3} \frac{x^4-81}{x-3}$$
= $$\ {Lt}_{x \rightarrow 3} \frac{4 x^3-0}{1}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= 4 × 33
= 108

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-1}{x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{n(1+x)^{n-1}}{n}$$
= n (1 + 0)n-1
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= n × 1n-1 = n.

Question 2.
(i) $$\underset{x \rightarrow 0}{\ {Lt}} \frac{\sin a x}{\sin b x}$$
(ii) $$\ {Lt}_{x \rightarrow 2} \frac{e^x-e^2}{x-2}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\ {Lt}_{x \rightarrow 0} \frac{a \cos a x}{b \cos b x}$$
= $$\frac{a \times 1}{b \times 1}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\frac{a}{b}$$

(ii) $$\ {Lt}_{x \rightarrow 2} \frac{e^x-e^2}{x-2}$$ = $$\ {Lt}_{x \rightarrow 2} \frac{e^x-0}{1-0}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= e2

Question 3.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x e^x}{1-e^x}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{\tan 2 x}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x e^x}{1-e^x}=\ {Lt}_{x \rightarrow 0} \frac{x e^x+e^x}{-e^x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{e^x(x+1)}{-e^x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ – (x + 1)
= – (0 + 1)
= – 1.

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{\tan 2 x}=\ {Lt}_{x \rightarrow 0} \frac{e^x}{2 \sec ^2 2 x}$$
= $$\frac{e^0}{2 \sec ^2 0}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\frac{1}{2 \times 1}=\frac{1}{2}$$

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

(ii) $$\ {Lt}_{x \rightarrow 1} \frac{x^2-x \log x+\log x-1}{x-1}$$ = $$\ {Lt}_{x \rightarrow 1} \frac{2 x-\left(x \times \frac{1}{x}+\log x \cdot 1\right)+\frac{1}{x}}{1}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\underset{x \rightarrow 1}{\mathrm{Lt}}$$ 2x – (1 + log x) + $$\frac{1}{x}$$
= 2 × 1 – (1 + log 1) + 1
= 2 – 1 – 0 + 1
= 2

Question 5.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{\cos x-1}{\cos 2 x-1}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{8^x-2^x}{4 x}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{\cos x-1}{\cos 2 x-1}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{-\sin x}{-2 \sin 2 x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{\sin x}{2 \sin 2 x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{\cos x}{4 \cos 2 x}$$
= $$\frac{1}{4 \times 1}=\frac{1}{4}$$

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{8^x-2^x}{4 x}$$
Solution:
$$\ {Lt}_{x \rightarrow 0} \frac{8^x-2^x}{4 x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{8^x \log 8-2^x \log 2}{4}$$
= $$\frac{8^0 \log 8-2^0 \log 2}{4}$$
[∵ log ab = b log a]
= $$\frac{\log 8-\log 2}{4}$$
= $$\frac{1}{4}$$ log 4
= $$\frac{1}{4}$$ × 2 log 2
= $$\frac{1}{2}$$ log 2

Question 6.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x-\tan x}{x-\sin x}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{2 \tan ^{-1} x-x}{2 x-\sin ^{-1} x}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x-\tan x}{x-\sin x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{1-\sec ^2 x}{1-\cos x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{-2 \sec ^2 x \tan x}{\sin x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{-2 \sec ^2 x \sin x}{\sin x \cos x}$$
= $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ – 2 sec3 x
= – 2 × 1 = – 2

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{2 \tan ^{-1} x-x}{2 x-\sin ^{-1} x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{\frac{2}{1+x^2}-1}{2-\frac{1}{\sqrt{1-x^2}}}$$
= $$\frac{2-1}{2-1}$$
= 1.

Question 7.
(i) $$\underset{x \rightarrow 0}{\ {Lt}} \frac{\log \sec 2 x}{\log \sec x}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{\cos 2 x-\cos x}{\sin ^2 x}$$
Solution:
(i) $$\underset{x \rightarrow 0}{\ {Lt}} \frac{\log \sec 2 x}{\log \sec x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{\sec 2 x} \sec 2 x \tan 2 x \cdot 2}{\frac{1}{\sec x} \sec x \tan x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{2 \tan 2 x}{\tan x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{4 \sec ^2 2 x}{\sec ^2 x}$$
= $$\frac{4 \times 1}{1}$$ = 4.

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

Question 8.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x-\sin x}{x^3}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x \sin x}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x-\sin x}{x^3}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{1-\cos x}{3 x^2}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{\sin x}{6 x}$$
= $$\frac{1}{6} \ {Lt}_{x \rightarrow 0} \frac{\sin x}{x}$$
= $$\frac{1}{6} \times 1=\frac{1}{6}$$

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x \sin x}$$

Question 9.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}+2 \cos x-4}{x^3}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{x^2+2 \cos x-2}{x \sin ^3 x}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}+2 \cos x-4}{x^3}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \sin x}{3 x^2}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-2 \cos x}{6 x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{e^x-o^{-x}+2 \sin x}{6}$$
= $$\frac{1-1+2 \times 0}{6}$$
= 0

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{x^2+2 \cos x-2}{x \sin ^3 x}$$

Question 10.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{\log (1-x)}{\tan \frac{\pi}{2} x}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{\left(\tan ^{-1} x\right)^2}{\log \left(1+x^2\right)}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{\log (1-x)}{\tan \frac{\pi}{2} x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{\frac{-1}{1-x}}{\frac{\pi}{2} \sec ^2 \frac{\pi}{2} x}$$
= $$\frac{\frac{-1}{1-0}}{\frac{\pi}{2} \times 1^2}$$
= – $$\frac{2}{\pi}$$

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{\left(\tan ^{-1} x\right)^2}{\log \left(1+x^2\right)}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{2 \tan ^{-1} x \cdot \frac{1}{1+x^2}}{\frac{2 x}{1+x^2}}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{\tan ^{-1} x}{x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{1+x^2}}{1}$$
= $$\frac{1}{1+0^2}$$ = 1.

Question 11.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x-\sin x \cos x}{x^3}$$ (ISC 2013)
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{\sin x}}{x-\sin x}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{x-\sin x \cos x}{x^3}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{x-\frac{1}{2} \sin 2 x}{x^3}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{1-\frac{1}{2} \cos 2 x \times 2}{3 x^2}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{2 \sin 2 x}{6 x}$$
= $$\frac{2}{3} \ {Lt}_{x \rightarrow 0} \frac{\sin 2 x}{2 x}$$
= $$\frac{2}{3} \times 1=\frac{2}{3}$$
[∵ $$\ {Lt}_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}$$ = 1]

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{\sin x}}{x-\sin x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]

Question 12.
(i) $$\ {Lt}_{x \rightarrow 0} \frac{\log \left(1-x^2\right)}{\log \cos x}$$
(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^3}$$
Solution:
(i) $$\ {Lt}_{x \rightarrow 0} \frac{\log \left(1-x^2\right)}{\log \cos x}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{\frac{-2 x}{1-x^2}}{-\frac{\sin x}{\cos x}}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{2 x}{\left(1-x^2\right) \tan x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{2}{1-x^2} \cdot \underset{x \rightarrow 0}{\ {Lt}} \frac{x}{\tan x}$$
= $$\frac{2}{1-0}$$ . 1 = 2
[∵ $$\ {Lt}_{\theta \rightarrow 0} \frac{\theta}{\tan \theta}$$ = 1]

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^3}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0} \frac{e^x \cos x+\sin x e^x-1-2 x}{3 x^2}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{e^x(-\sin x)+\cos x e^x+e^x \cos x+\sin x e^x-2}{6 x}$$
= $$\ {Lt}_{x \rightarrow 0} \frac{2 \cos x e^x-2}{6 x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{2 \cos x e^x-2 \sin x e^x}{6}$$
= $$\frac{2 \times 1 \times 1-2 \times 0 \times 1}{6}=\frac{1}{3}$$.

Question 13.
(i) $$\ {Lt}_{x \rightarrow 0^{+}} \frac{3^x-2^x}{\sqrt{x}}$$
(ii) $$\ {Lt}_{x \rightarrow 0^{+}} \frac{(1+x)^n-n x-1}{x^2}$$, n > 1
Solution:
(i) $$\ {Lt}_{x \rightarrow 0^{+}} \frac{3^x-2^x}{\sqrt{x}}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 0^{+}} \frac{3^x \log 3-2^x \log 2}{\frac{1}{2 \sqrt{x}}}$$
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ 2√x [3x log 3 – 2x log 2]
= 2 × 0 [1 × log 3 – 1 × log 2] = 0

(ii) $$\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-n x-1}{x^2}$$ ; n > 1
[Using L Hopital’s rule and ($$\frac{0}{0}$$ form)]
= $$\ {Lt}_{x \rightarrow 0} \frac{n(1+x)^{n-1}-n}{2 x}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{n(n-1)(1+x)^{n-2}-0}{2}$$
= $$\frac{n(n-1)}{2}$$

Question 14.
What is the fallacy in the following use of L’Hop[ital’s rule ?
$$\ {Lt}_{x \rightarrow 2} \frac{x^3-x^2-x-2}{x^3-3 x^2+3 x-2}$$ = $$\ {Lt}_{x \rightarrow 2} \frac{3 x^2-2 x-1}{3 x^2-6 x+3}$$ = $$\ {Lt}_{x \rightarrow 2} \frac{6 x-2}{6 x-6}=\ {Lt}_{x \rightarrow 2} \frac{6}{6}$$ = 1
Solution:
$$\ {Lt}_{x \rightarrow 2} \frac{x^3-x^2-x-2}{x^3-3 x^2+3 x-2}$$
[$$\frac{0}{0}$$ form, using ‘L’ Hopitals’s rule]
= $$\ {Lt}_{x \rightarrow 2} \frac{3 x^2-2 x-1}{3 x^2-6 x+3}$$
which is not ($$\frac{0}{0}$$) form so we can’t apply L’Hopital’s rule in this step.

Question 15.
If $$\underset{x \rightarrow 0}{\ {Lt}} \frac{\sin 2 x+k \sin x}{x^3}$$ is finite, find k and the limit.
Solution:
$$\ {Lt}_{x \rightarrow 0} \frac{\sin 2 x+k \sin x}{x^3}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x+k \cos x}{3 x^2}$$ ……….(1)
(using L’Hopital’s rule)
Since denominator of eqn. (1)
i.e. 3x2 → 0 as x → 0 so in order that given limit exist finitely.
Numerator of eqn. (1)
i.e. 2 cos 2x + k cos x → 0
as x → 0 and this happen.
When 2 × 1 + k × 1 = 0
⇒ k = – 2
= $$\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x-2 \cos x}{3 x^2}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{-4 \sin 2 x+2 \sin x}{6 x}$$ ($$\frac{0}{0}$$)
= $$\ {Lt}_{x \rightarrow 0} \frac{-8 \cos 2 x+2 \cos x}{6}$$
= $$\frac{-8 \times 1+2 \times 1}{6}$$
= $$\frac{-6}{6}$$ = – 1.

Question 16.
Find the values of a and b $$\ {Lt}_{x \rightarrow 0} \frac{x(1-a \cos x)+b \sin x}{x^3}$$ exists and equals $$\frac{1}{3}$$.
Solution:
$$\ {Lt}_{x \rightarrow 0} \frac{x(1-a \cos x)+b \sin x}{x^3}$$ ($$\frac{0}{0}$$ form)
= $$\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}$$ …………….(1)
Since denominator of eqn. (1) i.e. 3x2 → 0 as x → 0
so in order that given limit ecxists finitely, the numerator of eqn. (1)
i.e. (1 – a cos x) + ax sin x + b cos x → 0
as x → 0 and this happen
When 1 – a + b =0
Now eqn. (2) is satisfied.
When given limit
= $$\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}\left(\frac{0}{0} \text { form }\right)$$
= $$\ {Lt}_{x \rightarrow 0} \frac{a \sin x+a \sin x+a x \cos x-b \sin x}{6 x}\left(\frac{0}{0} \text { form }\right)$$
= $$\ {Lt}_{x \rightarrow 0} \frac{a \cos x+a \cos x+a \cos x-a x \sin x-b \cos x}{6}$$
= $$\frac{a+a+a-b}{6}$$
= $$\frac{3 a-b}{6}=\frac{1}{3}$$ (given)
⇒ 3a – b = 2
On solving (2) and (3) ; we have
1 + 2a = 2
⇒ a = $$\frac{1}{2}$$
and b = – $$\frac{1}{2}$$