We aim to evaluate the limit:

$\lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{1/x}$

Step-by-Step Solution:

1. Rewrite the expression inside the limit:
   Let: $L = \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{1/x}$ Taking the natural logarithm of both sides: $\ln L = \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{\tan x}{x} \right)$

2. Simplify $\ln \left( \frac{\tan x}{x} \right)$:
   $\ln \left( \frac{\tan x}{x} \right) = \ln (\tan x) - \ln (x)$ Therefore: $\ln L = \lim_{x \to 0} \frac{1}{x} \left[ \ln (\tan x) - \ln (x) \right]$

3. Expand $\ln (\tan x)$ around $x = 0$ using Taylor series:

   • For small $x$, $\tan x \approx x + \frac{x^3}{3}$, so: $\ln (\tan x) \approx \ln \left( x + \frac{x^3}{3} \right) \approx \ln x + \ln \left( 1 + \frac{x^2}{3} \right)$
   • Using $\ln (1 + u) \approx u$ for small $u$, we have: $\ln (\tan x) \approx \ln x + \frac{x^2}{3}$

4. Substitute into the expression for $\ln L$:
   $\ln (\tan x) - \ln (x) \approx \left( \ln x + \frac{x^2}{3} \right) - \ln x = \frac{x^2}{3}$ Thus: $\ln L = \lim_{x \to 0} \frac{1}{x} \cdot \frac{x^2}{3}$

5. Simplify the limit:
   $\ln L = \lim_{x \to 0} \frac{x}{3} = 0$

6. Exponentiate to find $L$:
   $L = e^{\ln L} = e^0 = 1$

Final Answer:

$\lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{1/x} = 1$

Would you like a detailed explanation of any step?

5 Related Questions:

1. What is the Taylor series expansion for $\tan x$?
2. How do you derive $\ln (1 + u) \approx u$ for small $u$?
3. What are some common techniques for evaluating logarithmic limits?
4. Can you solve similar problems with exponential limits?
5. What happens if the power $1/x$ is replaced with $x^2$?

Tip:

For logarithmic limits, always check for simplifications using series expansions or approximations to simplify complex expressions.