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JEE Mathematics 2025 Question with Solution

Evaluate the following limit: lim⁡x→0+tan⁡(5x13)log⁡(1+3x2)(tan⁡−1(3x))2(e5x43−1)\lim_{x \to 0^+} \frac{\tan\left(5x^{\frac{1}{3}}\right) \log\left(1 + 3x^2\right)}{\left(\tan^{-1}\left(3\sqrt{x}\right)\right)^2 \left(e^{5x^{\frac{4}{3}}} - 1\right)}

  • A

    115\frac{1}{15}

  • B

    11

  • C

    13\frac{1}{3}

  • D

    53\frac{5}{3}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

lim⁡x→0+tan⁡(5x13)log⁡(1+3x2)(tan⁡−1(3x))2(e5x43−1)\lim_{x \to 0^+} \frac{\tan\left(5x^{\frac{1}{3}}\right) \log\left(1 + 3x^2\right)}{\left(\tan^{-1}\left(3\sqrt{x}\right)\right)^2 \left(e^{5x^{\frac{4}{3}}} - 1\right)}

Find: The value of the limit.

For small values of xx, use the standard approximations:

tan⁡t≈t,tan⁡−1t≈t,log⁡(1+t)≈t,et−1≈t\tan t \approx t, \qquad \tan^{-1} t \approx t, \qquad \log(1+t) \approx t, \qquad e^t - 1 \approx t

Therefore,

tan⁡(5x13)≈5x13,log⁡(1+3x2)≈3x2,\tan\left(5x^{\frac{1}{3}}\right) \approx 5x^{\frac{1}{3}}, \qquad \log\left(1+3x^2\right) \approx 3x^2, tan⁡−1(3x)≈3x,e5x43−1≈5x43\tan^{-1}\left(3\sqrt{x}\right) \approx 3\sqrt{x}, \qquad e^{5x^{\frac{4}{3}}}-1 \approx 5x^{\frac{4}{3}}

Substituting these into the expression,

5x13⋅3x2(3x)2⋅5x43\frac{5x^{\frac{1}{3}} \cdot 3x^2}{\left(3\sqrt{x}\right)^2 \cdot 5x^{\frac{4}{3}}}

Now simplify:

15x739x⋅5x43\frac{15x^{\frac{7}{3}}}{9x \cdot 5x^{\frac{4}{3}}} =15x7345x73= \frac{15x^{\frac{7}{3}}}{45x^{\frac{7}{3}}} =1545=13= \frac{15}{45} = \frac{1}{3}

Therefore, the correct option is C and the value of the limit is 13\frac{1}{3}.

Factor-wise Limit Method

Given:

lim⁡x→0+tan⁡(5x13)log⁡(1+3x2)(tan⁡−1(3x))2(e5x43−1)\lim_{x \to 0^+} \frac{\tan\left(5x^{\frac{1}{3}}\right) \log\left(1 + 3x^2\right)}{\left(\tan^{-1}\left(3\sqrt{x}\right)\right)^2 \left(e^{5x^{\frac{4}{3}}} - 1\right)}

Find: The value of the limit.

Rewrite the expression as a product of standard limits:

(tan⁡(5x13)5x13)(log⁡(1+3x2)3x2)((3x)2(tan⁡−1(3x))2)(5x43e5x43−1)(5x13⋅3x2(3x)2⋅5x43)\left(\frac{\tan\left(5x^{\frac{1}{3}}\right)}{5x^{\frac{1}{3}}}\right) \left(\frac{\log\left(1+3x^2\right)}{3x^2}\right) \left(\frac{\left(3\sqrt{x}\right)^2}{\left(\tan^{-1}(3\sqrt{x})\right)^2}\right) \left(\frac{5x^{\frac{4}{3}}}{e^{5x^{\frac{4}{3}}}-1}\right) \left(\frac{5x^{\frac{1}{3}} \cdot 3x^2}{(3\sqrt{x})^2 \cdot 5x^{\frac{4}{3}}}\right)

As x→0+x \to 0^+, the first four factors tend to 11 using standard limits. The remaining algebraic factor is

15x739x⋅5x43=15x7345x73=13\frac{15x^{\frac{7}{3}}}{9x \cdot 5x^{\frac{4}{3}}} = \frac{15x^{\frac{7}{3}}}{45x^{\frac{7}{3}}} = \frac{1}{3}

Hence the whole limit is 13\frac{1}{3}.

This works because each transcendental term is isolated into a standard form whose limit is 11.

Common mistakes

  • Using tan⁡(5x1/3)≈x\tan(5x^{1/3}) \approx x instead of tan⁡(5x1/3)≈5x1/3\tan(5x^{1/3}) \approx 5x^{1/3}. This is wrong because the small-angle approximation preserves the full argument. Always replace tan⁡u\tan u by uu, not by xx.

  • Forgetting that (tan⁡−1(3x))2≈(3x)2=9x\left(\tan^{-1}(3\sqrt{x})\right)^2 \approx (3\sqrt{x})^2 = 9x. This is wrong because the entire inverse tangent term is squared. First approximate tan⁡−1(3x)\tan^{-1}(3\sqrt{x}), then square it.

  • Applying log⁡(1+t)≈t\log(1+t) \approx t and et−1≈te^t - 1 \approx t with the wrong inner term. Here the correct substitutions are t=3x2t = 3x^2 and t=5x4/3t = 5x^{4/3}. Use the complete inner expression each time.

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