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Maths Question 16 – JEE-MAIN 2025

limx→0+tan⁡(5x13)loge⁡(1+3x2)(tan−1⁡3x)2(e5x43−1) is equal to

This problem can be solved by applying standard limit approximations for functions like tan⁡u, loge⁡(1+u), tan−1⁡u, and eu−1 as u→0.

Step 1: Apply Standard Limit Approximations✦ Active

As x→0+, we use the following standard limit approximations for u→0: tan⁡u≈u, loge⁡(1+u)≈u, tan−1⁡u≈u, and eu−1≈u. Applying these to the terms in the given limit:

tan⁡(5x13)≈5x13 loge⁡(1+3x2)≈3x2 (tan−1⁡3x)2≈(3x)2=9x (e5x43−1)≈5x43
Step 2: Substitute Approximations into the Limit Expression○ Expand

Substitute these approximations back into the original limit expression:

L=limx→0+(5x13)(3x2)(9x)(5x43)
Step 3: Simplify and Evaluate the Limit○ Expand

Combine the terms in the numerator and denominator:

L=limx→0+15x13+245x1+43=limx→0+15x7345x73 Since x→0+,x≠0, we can cancel x73 terms: L=1545=13
💡 Teacher's Secret Hint

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

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