JEE Main 2025 Apr 7 Shift 1, Mathematics Q1: Methods to Evaluate Limits

JEE Main2025Apr 7, Shift 1Mathematics

$x \to 0^{+} lim \frac{tan ( 5 ( x ) ^{\frac{1}{3}} ) lo g _{e} ( 1 + 3 x ^{2} )}{( tan ^{- 1} 3 x ) ^{2} ( e ^{5 (x)^{\frac{4}{3}}} - 1 )}$ is equal to

A
$\frac{1}{15}$
B
$1$
C
$\frac{1}{3}$
D
$\frac{5}{3}$

Solution

Use the standard small-argument equivalences for $x \to 0^{+}$ :

$tan (5 x^{1/3}) \sim 5 x^{1/3}$
$lo g_{e} (1 + 3 x^{2}) \sim 3 x^{2}$
$tan^{- 1} (3 x) \sim 3 x$ , so its square $\sim 9 x$
$e^{5 x^{4/3}} - 1 \sim 5 x^{4/3}$

Substitute into the limit:

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

Group the powers of $x$ : numerator carries $x^{1/3 + 2} = x^{7/3}$ , denominator carries $x^{1 + 4/3} = x^{7/3}$ . The powers cancel.

$L = \frac{5 \cdot 3}{9 \cdot 5} = \frac{15}{45} = \frac{1}{3}$

$∴$ Option (3).

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q1: Methods to Evaluate Limits

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