Question

If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32

• A. 19117
• B. 18817
• C. 18280
• D. 19000

Hint:

In limit problems involving binomial expansions, use approximations for small values of \( x \) to simplify the expressions.

Answer 1

The Correct Option is C

We are given the limit: \[ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \] 

To solve for \( p \), we first use the fact that: \[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \] 

So the limit simplifies to: \[ \lim_{x \to 0} 1^{\frac{1}{x^2}} = 1 \] Thus, \( p = e^{\lim_{x \to 0} \frac{\ln \left( \frac{\tan x}{x} \right)}{x^2}} \). Using the approximation \( \frac{\tan x}{x} \approx 1 + \frac{x^2}{3} \), we find that \( p = e^{\frac{1}{3}} \). 

Therefore, the value of \( 96 \ln p \) is approximately 18280.