Answer: 32

Question

Answer: 32

Accepted Answer

To solve the given limit problem, we start by analyzing the expression:

$\underset{x	o 0}{lim}{(\frac{	an x}{x})}^{\frac{1}{{x}^{2}}}=p$

This limit exhibits the indeterminate form ${1}^{\infty}$. To handle this form, we use the transformation:

$p={e}^{\underset{x	o 0}{lim}(\frac{	an x}{x}-1)\frac{1}{{x}^{2}}}$

Expanding $	an x$ using its Taylor series near $x=0$, we have:

$	an x=x+\frac{{x}^{3}}{3}+\frac{2}{15}{x}^{5}+\ldots$

Substituting this expansion into the limit, we get:

$\frac{	an x-x}{{x}^{3}}=\frac{(x+\frac{{x}^{3}}{3}+\frac{2}{15}{x}^{5}+\ldots -x)}{{x}^{3}}=\frac{\frac{{x}^{3}}{3}+\frac{2}{15}{x}^{5}+\ldots }{{x}^{3}}$

This simplifies to:

$\frac{{x}^{3}}{3{x}^{3}}=\frac{1}{3}$

Thus, the limit becomes:

$p={e}^{\frac{1}{3}}$

Therefore, the expression for ${\log }_{e}p$ is:

${\log }_{e}p=\frac{1}{3}$

Finally, computing $96{\log }_{e}p$:

$96{\log }_{e}p=96\cdot \frac{1}{3}=32$

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