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MHTCET 20 Apr 2026 Shift 1 Paper

Question : 112 of 150

Marks: +1, -0

\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+4 x)}{\left(e^{4 x}-1\right)^{2}}\right]=\cdots

Solution:

Concept:

Use the standard limits

\lim_{t \to 0} \frac{\log(1+t)}{t} = 1

and

\lim_{t \to 0} \frac{e^t - 1}{t} = 1

Explanation:

Let the given limit be

L

Rewrite the expression to apply the standard forms:

\frac{x \log(1+4x)}{(e^{4x}-1)^2} = \frac{x(4x)\frac{\log(1+4x)}{4x}}{(4x)^2\left(\frac{e^{4x}-1}{4x}\right)^2}

Simplify the coefficient:

\frac{4x^2}{16x^2} = \frac{1}{4}

So,

L = \lim_{x \to 0} \frac{1}{4} \cdot \frac{\frac{\log(1+4x)}{4x}}{\left(\frac{e^{4x}-1}{4x}\right)^2}

x \to 0

, we get

\frac{\log(1+4x)}{4x} \to 1

and

\frac{e^{4x}-1}{4x} \to 1

Therefore,

L = \frac{1}{4} \cdot \frac{1}{1^2} = \frac{1}{4}

Answer:

The limit is

\frac{1}{4}

, which corresponds to Option A.

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