If operatorname{Lim}_{x rightarrow 0}left(fra | vedclass

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(C) Let $P = \lim_{x ightarrow 0} \left(\frac{	an x}{x}ight)^{\frac{1}{x^2}}$.Since the form is $1^{\infty}$,we use the formula $\lim_{x ightar

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Let $P = \lim_{x ightarrow 0} \left(\frac{	an x}{x}ight)^{\frac{1}{x^2}}$.Since the form is $1^{\infty}$,we use the formula $\lim_{x ightarrow a} f(x)^{g(x)} = e^{\lim_{x ightarrow a} (f(x)-1)g(x)}$.$P = e^{\lim_{x ightarrow 0} \left(\frac{	an x}{x} - 1ight) \frac{1}{x^2}} = e^{\lim_{x ightarrow 0} \frac{	an x - x}{x^3}}$.Using the Taylor series expansion $	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$,we get:$P = e^{\lim_{x ightarrow 0} \frac{(x + \frac{x^3}{3} + \dots) - x}{x^3}} = e^{\lim_{x ightarrow 0} \frac{x^3/3}{x^3}} = e^{1/3}$.Thus,$\log_e p = \log_e (e^{1/3}) = \frac{1}{3}$.Therefore,$96 \log_e p = 96 	imes \frac{1}{3} = 32$.

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

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