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MHTCET 10 May 2024 Shift 2 Paper

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Question : 129 of 150

Marks: +1, -0

If

\mathrm{f}(x)=\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}

is continuous at

x=0

f(0)

is equal to

0
1
e
$\frac{1}{e}$

Solution:

Concept:

This limit is of the form

1^{\infty}

, so take the natural logarithm and apply standard small-angle approximations.

Explanation:

Since

f

is continuous at

x=0

,

f(0)=\lim_{x\to 0} f(x)

.

Let

L=\lim_{x\to0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}

.

Taking

\ln

on both sides:

\ln L=\lim_{x\to0}\operatorname{cosec} x \cdot \ln\left(\frac{1+\tan x}{1+\sin x}\right)

.

This becomes:

\ln L=\lim_{x\to0}\frac{\ln(1+\tan x)-\ln(1+\sin x)}{\sin x}

.

Using

\ln(1+u)=u-\frac{u^2}{2}+\cdots

, higher-order terms vanish as

x\to0

:

\ln L=\lim_{x\to0}\frac{\tan x-\sin x}{\sin x}

.

Simplify:

\frac{\tan x-\sin x}{\sin x}=\frac{1}{\cos x}-1

.

Therefore:

\ln L=\lim_{x\to0}\left(\frac{1}{\cos x}-1\right)=1-1=0

.

So

L=e^{0}=1

.

Hence

f(0)=1

.

Answer:

f(0)=1

, which is option B.

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