Question

Let \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) + \tan^3 \left( \frac{x}{2^{r+1}} \right)}{1 - \tan^2 \left( \frac{x}{2^{r+1}} \right)} \right) \] Then, \( \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} \) is equal to:

• A. 1
• B. 0
• C. \( \infty \)
• D. \( -1 \)

Hint:

For limits involving exponential functions and trigonometric functions, consider using L'Hopital's Rule to simplify the expression when the limit results in an indeterminate form.

Answer 1

The Correct Option is A

We are given that: \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) - \tan \left( \frac{x}{2^{r+2}} \right)}{1} \right) \] This simplifies to: \[ f(x) = \tan x \] Next, calculate the limit: \[ \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} = \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} \] Using L'Hopital's Rule, we compute: \[ \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} = 1 \]