If $\underset{\mathrm{x} \rightarrow 0}{\operatorname{Lim}}\left(\frac{\tan \mathrm{x}}{\mathrm{x}}\right)^{\frac{1}{\mathrm{x}^{2}}}=\mathrm{p}$, then $96 \log _{\mathrm{e}} \mathrm{p}$ is equal to

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Limits

x \to 0 Lim (\frac{t a n x}{x})^{\frac{1}{x ^{2}}} = p

, then

96 lo g_{e} p

is equal to

Answer: 32

Write

p = lim_{x \to 0} (\frac{t a n x}{x})^{1/ x^{2}} = exp (lim_{x \to 0} \frac{l n ( t a n x / x )}{x ^{2}})

. Since

tan x = x + \frac{x ^{3}}{3} + O (x^{5})

\frac{t a n x}{x} = 1 + \frac{x ^{2}}{3} + O (x^{4})

and

ln (\frac{t a n x}{x}) = \frac{x ^{2}}{3} + O (x^{4})

, so the exponent tends to

\frac{1}{3}

and

p = e^{1/3}

. Therefore

96 ln p = 96 \cdot \frac{1}{3}

, which equals

32

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Integer answer

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