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JEE MAIN 2025
07-04-2025 SHIFT-1
Question
$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\tan \left( {5{{(x)}^{\frac{1}{3}}}} \right){{\log }_e}\left( {1 + 3{x^2}} \right)}}{{{{\left( {{{\tan }^{ - 1}}3\sqrt x } \right)}^2}\left( {{e^{5{{(x)}^{\frac{4}{3}}}}} - 1} \right)}}$ is equal to
Select the correct option:
A
$\frac{5}{3}$
B
$\frac{1}{{15}}$
C
1
D
$\frac{1}{3}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{\tan \left( {5{x^{1/3}}} \right)}}{{5{x^{1/3}}}}} \right) \cdot \left( {\frac{{{{(3\sqrt x )}^2}}}{{{{\left( {{{\tan }^{ - 1}}3\sqrt x } \right)}^2}}}} \right)\left( {\frac{{\ell \left( {1 + 3{x^2}} \right)}}{{3{x^2}}}} \right)\left( {\frac{{5{x^{4/3}}}}{{{e^{5{x^{\frac{4}{3}}}}} - 1}}} \right) \times \frac{{5{x^{1/3}} \cdot 3{x^2}}}{{5{x^{4/3}} \cdot 9x}}$ $ = \frac{1}{3}$
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