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

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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 × 1^n-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 a^b = 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 [3^x log 3 – 2^x 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 = – \(\frac{1}{2}\)