Test Index

MHTCET 13 Apr 2026 Shift 1 Paper

© examsnet.com

Question : 110 of 150

Marks: +1, -0

\lim _{x \rightarrow 0}\left[\frac{\log |2+x|-\log |2-x|}{\tan x}\right]=\ldots \ldots

.

2
0
-1
1

Solution:

Concept:

The limit is of the form

\frac{0}{0}

, so L'Hôpital's rule can be applied.

Explanation:

For

x

close to

0

,

2+x

and

2-x

are positive.

Therefore,

\log|2+x| = \log(2+x)

and

\log|2-x| = \log(2-x)

.

The given limit becomes

\lim_{x \to 0} \frac{\log(2+x)-\log(2-x)}{\tan x}

.

Using logarithm properties:

\log(2+x)-\log(2-x) = \log\left(\frac{2+x}{2-x}\right)

.

Differentiating numerator:

\frac{d}{dx}[\log(2+x)-\log(2-x)] = \frac{1}{2+x}+\frac{1}{2-x}

.

Differentiating denominator:

\frac{d}{dx}(\tan x) = \sec^2 x

.

So, the limit becomes

\lim_{x \to 0} \frac{\frac{1}{2+x}+\frac{1}{2-x}}{\sec^2 x}

.

Substituting

x=0

gives

\frac{\frac{1}{2}+\frac{1}{2}}{1} = 1

.

Answer:

1

Correct option: D

© examsnet.com

Go to Question:

Prev Question

Next Question