If _x0(( x)/(x))^1/x^2=p, then 96_ep is equal to____________ Answer: 32

Question

If $\displaystyle\lim_{x	o0}\left(\frac{	an x}{x}ight)^{1/x^2}=p$, then $96\log_ep$ is equal to\_\_\_\_\_\_\_\_\_\_\_\_

Answer: 32

Accepted Answer

This is a $1^\infty$ form. Taking logs, $\ln p=\displaystyle\lim_{x\to0}\frac{1}{x^2}\left(\frac{\tan x}{x}-1\right)=\lim_{x\to0}\frac{\tan x-x}{x^3}$.

Using $\tan x=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\cdots$:
$$\tan x-x=\frac{x^3}{3}+\frac{2x^5}{15}+\cdots \implies \frac{\tan x-x}{x^3}\to\frac13$$

So $\ln p=\dfrac13$, and $96\log_ep=96\cdot\dfrac13=32$.

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