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Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 7 April, Morning Shift — Question 37

lim⁡x→0+tan⁡(5(x)13)log⁡e(1+3x2)(tan⁡−13x)2(e5(x)43−1)\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

  1. Option A:

    115\frac{1}{15}

  2. Option B:

    53\frac{5}{3}

  3. Option C:

    13\frac{1}{3}

    Correct
  4. Option D:

    1

Answer: C

Step-by-step solution

lim⁡x→0+tan⁡(5(x)13)ln⁡(1+3x3)(tan⁡−1(3x))2(e5x43−1)\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)}

tan⁡(5(x)13)5(x)13ln⁡(1+3x2)3x2×5(x)13(3x2)(tan⁡−1(3x))2(3x)2(e5x43−1)5x13×9x×5x43\frac{\frac{\tan \left(5(x)^{\frac{1}{3}}\right)}{5(x)^{\frac{1}{3}}} \frac{\ln \left(1+3 x^{2}\right)}{3 x^{2}} \times 5(x)^{\frac{1}{3}}\left(3 x^{2}\right)}{\frac{\left(\tan ^{-1}(3 \sqrt{x})\right)^{2}}{(3 \sqrt{x})^{2}} \frac{\left(e^{5 x^{\frac{4}{3}}}-1\right)}{5 x^{\frac{1}{3}}} \times 9 x \times 5 x^{\frac{4}{3}}} lim⁡x→0+tan⁡(5(x)13)5(x)13ln⁡(1+3x3)3x2×15x73tan⁡−1(3x))2(3x)2(e5(x)43−1)5x13×45x73=13\begin{aligned} & \lim _{x \rightarrow 0^{+}} \frac{\frac{\tan \left(5(x)^{\frac{1}{3}}\right)}{5(x)^{\frac{1}{3}}} \frac{\ln \left(1+3 x^{3}\right)}{3 x^{2}} \times 15 x^{\frac{7}{3}}}{\frac{\left.\tan ^{-1}(3 \sqrt{x})\right)^{2}}{(3 \sqrt{x})^{2}} \frac{\left(e^{5(x)^{\frac{4}{3}}}-1\right)}{5 x^{\frac{1}{3}}} \times 45 x^{\frac{7}{3}}} \\& \quad=\frac{1}{3} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
7 April, Morning Shift
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Evaluation of Limit of Functions

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