ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.4

Well-structured ML Aggarwal Class 12 Solutions Chapter 6 Indeterminate Forms Ex 6.4 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.4

Typical Problems:

Question 1.
(i) Evaluate the following limits:
(i) \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) x tan^-1 (\(\frac{2}{x}\))
(ii) \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) \(\frac{\sin ^{-1} \frac{1}{x}}{\tan \frac{1}{x}}\)
Solution:
(i) Put \(\frac{1}{x}\) = t
as x → ∞ ⇒ t → 0
∴ \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) x tan^-1 (\(\frac{2}{x}\)) = \(\underset{t \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\tan ^{-1}(2 t)}{t}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{t \rightarrow 0} \frac{2}{1+4 t^2}\)
= \(\frac{2}{1+4 \times 0}\) = 2.

(ii) Put \(\frac{1}{x}\) = t
⇒ x = \(\frac{1}{t}\)
as x → ∞ ⇒ t → 0
∴ \(\ {Lt}_{x \rightarrow \infty} \frac{\sin ^{-1} \frac{1}{x}}{\tan \frac{1}{x}}=\underset{t \rightarrow 0}{\mathrm{Lt}} \frac{\sin ^{-1} t}{\tan t}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{t \rightarrow 0} \frac{\frac{1}{\sqrt{1-t^2}}}{\sec ^2 t}\)
= \(\ {Lt}_{t \rightarrow 0} \frac{1}{\sec ^2 t \sqrt{1-t^2}}\)
= \(\frac{1}{(1)^2 \times \sqrt{1-0^2}}\) = 1.

Question 2.
If \(\frac{\sin 2 x+k \sin x}{x^3}\) is finite, find k and the limit.
Solution:
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\sin 2 x+k \sin x}{x^3}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x+k \cos x}{3 x^2}\) …………(1)
Now deno 3x² → 0 as x → 0 and limit is given to be finite.
So it is necessary that Numerator of eqn. (1)
i.e. 2 cos 2x + k cos x vanishes as x → 0
and this happens when 2 + k = 0
⇒ k = – 2
∴ From (1) ; we have
given limit = \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x-2 \cos x}{3 x^2}\) ((\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{-4 \sin 2 x+2 \sin x}{6 x}\) ((\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{-8 \cos 2 x+2 \cos x}{6}\)
= \(\frac{-8 \times 1+2 \times 1}{6}\) = – 1.

Question 3.
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, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}\) ………….(1)
Now the denominator of eqn. (1)
i.e. 3x² → 0 as x → 0.
∴ the given limit exists finitely, it is necessary that Numerator of eqn. (1) is also goes to 0 as x → 0 and this happen when
1 – a × 1 + a × 0 + b = 0
⇒ 1 – a + b = 0
⇒ a – b = 1 ………..(2)
Let eqn. (2) happens, Then given limit
= \(\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a \sin x+a \sin x+a x \cos x-b \sin x}{6 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 a \sin x+a x \cos x-b \sin x}{6 x}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 a \cos x+a \cos x-a x \sin x-b \cos x}{6}\)
= \(\frac{2 a+a-b}{6}\)
Also given limit = \(\frac{1}{3}\)
∴ \(\frac{2 a+a-b}{6}\) = \(\frac{1}{3}\)
⇒ 3a – b = 2 ………..(3)
On solving eqn. (2) and eqn. (3) ; we have
a = \(\frac{1}{2}\)
and b = – \(\frac{1}{2}\).

Question 4.
Find the values of a, b and c so that \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x \sin x}\) = 2.
Solution:
\(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x \sin x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x^2}\left(\frac{x}{\sin x}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x^2}\) . 1 ……….(1)
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\theta}{\sin \theta}\) = 1]
Here denominator of eqn. (1) = x² → 0 as x → 0.
Now the limit exists finitely if Numerator of eqn. (1)
i.e. ae^x – b cos x + c e^-x as x → 0
and thos happen if a – b + c = 0 ……….(2)
suppose eqn.(2) is holds. Then given limit
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x+b \sin x-c e^{-x}}{2 x}\) ………..(3)
Now denominator of eqn. (3) i.e. 2x → 0 as x → 0
∴ in order that the limit exists finitely it is necessary that Numerator of eqn. (3) vanishes as x → 0.
This happens if a × 1 + b × 0 – c = 0
⇒ a – c = 0 ………..(4)
Now eqn. (4) holds.
Then given limit = \(\ {Lt}_{x \rightarrow 0} \frac{a e^x+b \sin x-c e^{-x}}{2 x}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x+b \cos x+c e^{-x}}{2}\)
= \(\frac{a+b+c}{2}\)
Also given limit = 2
∴ a + b + c = 4 ……..(5)
from (4) ;
a = c
∴ from (1) ;
2c – b = 0 …………..(6)
and from (5) ;
b + 2c = 4 …………..(7)
On solving eqn. (6) and (7) ; we have
c = 1; a = 1; b = 2.

Question 5.
Evaluate the following limits :
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log _a x}{x^k}\), k > 0
(ii) \(\ {Lt}_{x \rightarrow \infty} 2^x \sin \frac{a}{2^x}\), a ≠ 0
Solution:
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log _a x}{x^k}\), k > 0
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{x} \frac{1}{\log a}}{k x^{k-1}}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(\log a) k x^k}\), k > 0 = 0

(ii) put \(\frac{a}{2^x}\) = t
as x → ∞
⇒ 2^x → ∞
⇒ \(\frac{a}{2^x}\) → 0
⇒ t → 0 [a ≠ 0]
∴ \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) 2^x sin \(\frac{a}{2^x}\)
= \(\underset{t \rightarrow 0}{\mathrm{Lt}}\) \(\frac{a}{t}\) sin t
= a \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) \(\frac{sin t}{t}\)
= a × 1 = a.

Question 6.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x}\)
(ii) \(\ {Lt}_{x \rightarrow a}\left(2-\frac{a}{x}\right)^{\tan \frac{\pi x}{2 a}}\).
Solution:
(i) Let F(x) = \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x}\)
⇒ log F(x) = \(\frac{1}{x}\) log \(\left(\frac{\tan x}{x}\right)\)

(ii) Let F(x) = \(\left(2-\frac{a}{x}\right)^{\tan \frac{\pi x}{2 a}}\)

Question 7.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x^2}\)
(ii) \(\ {Lt}_{x \rightarrow 0}(1+2 x)^{\frac{x+5}{x}}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x^2}\)

∴ log (\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x)) = \(\frac{1}{3}\)
⇒ F(x) = e^1/3

(ii) Let F(x) = \(\ {Lt}_{x \rightarrow 0}(1+2 x)^{\frac{x+5}{x}}\)
⇒ log F(x) = \(\left(\frac{x+5}{x}\right)\) log (1 + 2x)
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\left(\frac{x+5}{x}\right)\) log (1 + 2x)
= \(\ {Lt}_{x \rightarrow 0} \frac{(x+5) \log (1+2 x)}{x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{(x+5) 2}{1+2 x}+\log (1+2 x)}{1}\)
= \(\frac{(0+5) 2}{1+0}\) + log (1 + 0) = 0
∴ log \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x) = 10
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\ {Lt}_{x \rightarrow 0}(1+2 x)^{\frac{x+5}{x}}\) = e¹⁰

Question 8.
Evaluate : \(\ {Lt}_{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\)
Solution:
\(\ {Lt}_{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\) = \(\ {Lt}_{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\frac{\sin x}{x}}{1-\frac{\sin x}{x}}}\)