lim_{x -> 0^+} tan(5(x)^(1/3)) log_e(1+3x^2) / (tan^-1 3sqrt(x))^2 (e^(5(x)^(4/3))-1) is equal to

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2025

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Home Maths Limits and DerivativesQuestion #277

lim_{x \to 0^{+}} \frac{\tan (5 x^{\frac{1}{3}}) \log_{e} (1 + 3 x^{2})}{(\tan^{- 1} 3 \sqrt{x})^{2} (e^{5 x^{\frac{4}{3}}} - 1)}

is equal to

Clue 1 of 3: Standard Limits

This problem can be solved by applying standard limit approximations for functions like $\tan u$ , $\log_{e} (1 + u)$ , $\tan^{- 1} u$ , and $e^{u} - 1$ as $u \to 0$ .

🧠 Full Solution Path

Step 1: Apply Standard Limit Approximations✦ Active

As $x \to 0^{+}$ , we use the following standard limit approximations for $u \to 0$ : $\tan u \approx u$ , $\log_{e} (1 + u) \approx u$ , $\tan^{- 1} u \approx u$ , and $e^{u} - 1 \approx u$ . Applying these to the terms in the given limit:

\tan (5 x^{\frac{1}{3}}) \approx 5 x^{\frac{1}{3}}

\log_{e} (1 + 3 x^{2}) \approx 3 x^{2}

(\tan^{- 1} 3 \sqrt{x})^{2} \approx (3 \sqrt{x})^{2} = 9 x

(e^{5 x^{\frac{4}{3}}} - 1) \approx 5 x^{\frac{4}{3}}

Step 2: Substitute Approximations into the Limit Expression○ Expand

Substitute these approximations back into the original limit expression:

L = lim_{x \to 0^{+}} \frac{(5 x^{\frac{1}{3}}) (3 x^{2})}{(9 x) (5 x^{\frac{4}{3}})}

Step 3: Simplify and Evaluate the Limit○ Expand

Combine the terms in the numerator and denominator:

L = lim_{x \to 0^{+}} \frac{15 x^{\frac{1}{3} + 2}}{45 x^{1 + \frac{4}{3}}} = lim_{x \to 0^{+}} \frac{15 x^{\frac{7}{3}}}{45 x^{\frac{7}{3}}}

Since x \to 0^{+}, x \neq 0, we can cancel x^{\frac{7}{3}} terms:

L = \frac{15}{45} = \frac{1}{3}

💡 Teacher's Secret Hint

Ensure correct addition of exponents: $1 / 3 + 2 = 7 / 3$ and $1 + 4 / 3 = 7 / 3$ .

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Maths Question 16 – JEE-MAIN 2025

Clue 1 of 3: Standard Limits

🧠 Full Solution Path