Medium MCQ +4 / -1 PYQ · JEE Mains 2025

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

  1. A $\frac{5}{3}$
  2. B 1
  3. C $\frac{1}{3}$ Correct answer
  4. D $\frac{1}{15}$

Solution

<p>$\lim _{x \rightarrow 0^{+}} \frac{\tan \left(5(x)^{\frac{1}{3}}\right) \ln \left(1+3 x^3\right)}{\left(\tan ^{-1}(3 \sqrt{x})\right)^2\left(e^{5 x^{\frac{4}{3}}}-1\right)}$</p> <p>$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^2})} \over {3{x^2}}} \times 5{{(x)}^{{1 \over 3}}}(3{x^2})} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5x{4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 9x \times 5{x^{{4 \over 3}}}}}$$</p> <p>$$\mathop {\lim }\limits_{x \to {0^ + }} {{{{\tan \left( {5{{(x)}^{{1 \over 3}}}} \right)} \over {5{{(x)}^{{1 \over 3}}}}}{{\ln (1 + 3{x^3})} \over {3{x^2}}} \times 15{x^{{7 \over 3}}}} \over {{{{{\left( {{{\tan }^2}\left( {3\sqrt x } \right)} \right)}^2}} \over {{{\left( {3\sqrt x } \right)}^2}}}{{\left( {{e^{5(x){4 \over 3}}} - 1} \right)} \over {5{x^{{1 \over 3}}}}} \times 45{x^{{7 \over 3}}}}}$$</p> <p>$= {1 \over 3}$</p>

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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