_ x arrow 0^ + (5(x)^ (1)/(3) ) _e(1+3 x^2) ( ^ -1 3 x )^2(e^ 5(x)^ (4)/(3) -1) is equal to

Correct answer: C. (1)/(3) aligned & _ x arrow 0^ · ( (5 x^ 1 / 3 ) 5 x^ 1 / 3 ) ·( (3 x )^2 ( ^ -1 3 x )^2 )( (1+3 x^2) 3 x^2 )( 5 x^ 4 / 3 e^ 5 x^4 -1 ) × 5 x^ 1 / 3 · 3 x^2 5 x^ 4 / 3 · 9 x & =(1)/(3) aligned

JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Limits

_ x arrow 0^ + (5(x)^ (1)/(3) ) _e(1+3 x^2) ( ^ -1 3 x )^2(e^ 5(x)^...

\lim _{x \rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \log _e\left(1+3 x^2\right)}{\left(\tan ^{-1} 3 \sqrt{x}\right)^2\left(e^{5(x)^{\frac{4}{3}}}-1\right)}

is equal to

Options

A $\frac{1}{15}$
B1
C $\frac{1}{3}$
D $\frac{5}{3}$

Correct answer

C. $\frac{1}{3}$

Step-by-step solution

\begin{aligned} & \lim _{x \rightarrow 0^{\cdot}}\left(\frac{\tan \left(5 \mathrm{x}^{1 / 3}\right)}{5 \mathrm{x}^{1 / 3}}\right) \cdot\left(\frac{(3 \sqrt{\mathrm{x}})^2}{\left(\tan ^{-1} 3 \sqrt{\mathrm{x}}\right)^2}\right)\left(\frac{\ell\left(1+3 \mathrm{x}^2\right)}{3 \mathrm{x}^2}\right)\left(\frac{5 \mathrm{x}^{4 / 3}}{\mathrm{e}^{5 x^4}-1}\right) \times \frac{5 \mathrm{x}^{1 / 3} \cdot 3 \mathrm{x}^2}{5 \mathrm{x}^{4 / 3} \cdot 9 \mathrm{x}} \\ & =\frac{1}{3}\end{aligned}

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