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MHTCET 23 Apr 2025 Shift 2 Paper

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Question : 107 of 150

Marks: +1, -0

Let

A=\lim _{x \rightarrow 0^{+}}\left(1+\tan ^{2} \sqrt{x}\right)^{\frac{1}{2 x}}

\log _{e} A=

2
1
$\frac{1}{2}$
$\frac{1}{4}$

Solution:

Concept:

Use the standard limit form for

1^{\infty}

: if

f(x) \to 1

and

g(x) \to \infty

, then

\lim [f(x)]^{g(x)} = e^{\lim g(x)(f(x)-1)}

.

Explanation:

Given

A=\lim_{x \to 0^{+}} \left(1+\tan^2 \sqrt{x}\right)^{\frac{1}{2x}}

.

Here

f(x)=1+\tan^2 \sqrt{x} \to 1

and

g(x)=\frac{1}{2x} \to \infty

.

So,

A = e^{\lim_{x \to 0^{+}} \frac{1}{2x}\left(\tan^2 \sqrt{x}\right)}

.

Rewrite it as

A = e^{\lim_{x \to 0^{+}} \frac{1}{2} \left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)^{2}}

.

Using

\lim_{x \to 0^{+}} \frac{\tan \sqrt{x}}{\sqrt{x}} = 1

, we get

A = e^{\frac{1}{2}}

.

Therefore,

\log_{e} A = \log_{e} e^{1/2} = \frac{1}{2}

.

Answer:

\log_{e} A = \frac{1}{2}

, which matches option C.

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