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

Step 1: Function Analysis

The provided function is: \[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan \left( \frac{x}{2^{r+1}} \right) - \tan \left( \frac{x}{2^{r+2}} \right)}{1} \right) \] This simplifies to: \[ f(x) = \tan x \]

Step 2: Limit Setup

The limit to be calculated is: \[ \lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)} = \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} \]

Step 3: L'Hopital's Rule Application

As this is an indeterminate form $\frac{0}{0}$, L'Hopital's Rule is applied. The numerator and denominator are differentiated: \[ \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} \] Derivative of the numerator: \[ \frac{d}{dx} \left( e^x - e^{\tan x} \right) = e^x - e^{\tan x} \cdot \sec^2 x \] Derivative of the denominator: \[ \frac{d}{dx} (x - \tan x) = 1 - \sec^2 x \] Evaluation at $x = 0$: \[ \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x} = 1 \]

Final Answer: 1