$$If\,\,\mathop {\lim }\limits_{x \to 0} \left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}=p \text {, then } 96 \log _{\mathrm{e}} p \text { is equal to____________ }$$

<p>To solve the given limit problem, we start by analyzing the expression:</p> <p>$ \lim\limits_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = p $</p> <p>This limit exhibits the indeterminate form $1^{\infty}$. To handle this form, we use the transformation:</p> <p>$ p = e^{\lim_{x \to 0} \left( \frac{\tan x}{x} - 1 \right) \frac{1}{x^2}} $</p> <p>Expanding $\tan x$ using its Taylor series near $x = 0$, we have:</p> <p>$ \tan x = x + \frac{x^3}{3} + \frac{2}{15}x^5 + \ldots $</p> <p>Substituting this expansion into the limit, we get:</p> <p>$ \frac{\tan x - x}{x^3} = \frac{\left(x + \frac{x^3}{3} + \frac{2}{15}x^5 + \ldots - x\right)}{x^3} = \frac{\frac{x^3}{3} + \frac{2}{15}x^5 + \ldots}{x^3} $</p> <p>This simplifies to:</p> <p>$ \frac{x^3}{3x^3} = \frac{1}{3} $</p> <p>Thus, the limit becomes:</p> <p>$ p = e^{\frac{1}{3}} $</p> <p>Therefore, the expression for $\log_e p$ is:</p> <p>$ \log_e p = \frac{1}{3} $</p> <p>Finally, computing $96 \log_e p$:</p> <p>$ 96 \log_e p = 96 \cdot \frac{1}{3} = 32 $</p>