ₓ arrow 0} {≤ft(5^{x}-1)^{4} {cosec}(x 5)}{ (x 5) · ≤ft(1+x² 25)}=

Correct Answer is : log⁡5 {5}log5

Correct Answer is : log⁡5 {5}log5

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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 110 of 150

Marks: +1, -0

\lim _{x \rightarrow 0} \frac{\left(5^{x}-1\right)^{4} \operatorname{cosec}(x \log 5)}{\tan (x \log 5) \cdot \log \left(1+x^{2} \log 25\right)}=\ldots \ldots

Solution:

Concept:

Use standard limit forms near

x \to 0

\frac{a^x - 1}{x} \to \log a

\frac{\sin \theta}{\theta} \to 1

\frac{\tan \theta}{\theta} \to 1

, and

\frac{\log(1+u)}{u} \to 1

Explanation:

x \to 0

, replace each factor by its equivalent small quantity.

5^x - 1 \sim x \log 5

\csc(x \log 5) \sim \frac{1}{x \log 5}

\tan(x \log 5) \sim x \log 5

\log\left(1 + x^2 \log 25\right) \sim x^2 \log 25 = 2x^2 \log 5

Substitute these into the given limit.

L = \lim_{x \to 0} \frac{(x \log 5)^4 \cdot \frac{1}{x \log 5}}{(x \log 5)(2x^2 \log 5)}

Simplify the powers of

x

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

L = \frac{(\log 5)^4}{(\log 5)^2 \cdot 2 \log 5} = \frac{\log 5}{2}

Since

\frac{\log 5}{2} = \log(5^{1/2}) = \log \sqrt{5}

Answer:

The correct option is B:

\log \sqrt{5}

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