If lim x 0 ( tan x x ) 1 x 2 = p , then 96 log e p is equal to __________. [2025]

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Solution

Q.
If $\lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x^{2}}} = p$ , then $96 \log_{e} p$ is equal to __________. [2025]

Ans.
(32)

Given, $p = \lim_{x \to 0} {(\frac{\tan x}{x})}^{\frac{1}{x^{2}}}$

$\Rightarrow p = e^{\lim_{x \to 0} (\frac{\tan x - x}{x^{3}})}$

$= e^{\lim_{x \to 0} (\frac{x + \frac{x^{3}}{3} + \frac{2 x^{5}}{5} + . . . - x}{x^{3}})}$

$[∵ For 1^{\infty} form limits, \lim_{x \to a} f {(x)}^{g (x)} = e^{\lim_{x \to a} (f (x) - 1) g (x)}]$

$= e^{1 / 3}$

$∴ 96 \log_{e} p = 96 \times \frac{1}{3} = 32$

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