Let $$f(x) = \lim_{n \to \infty} \sum_{r=0}^{n}\left(\frac{\tan\left(\frac{x}{2^{r+1}}\right) + \tan^{3}\left(\frac{x}{2^{r+1}}\right)}{1 - \tan^{2}\left(\frac{x}{2^{r+1}}\right)}\right).\quad\text{Then }\lim_{x \to 0} \frac{e^{x} - e^{f(x)}}{x - f(x)}$$ is equal to.

We begin by simplifying the general term in the sum. Let $$a = \frac{x}{2^{r+1}}.$$ We use the double‐angle identity for tangent:

Formula: $$\tan(2a) = \frac{2\tan a}{1 - \tan^2 a}.$$

From this, we get

$$\tan(2a) - \tan a = \frac{2\tan a}{1 - \tan^2 a} - \tan a = \frac{2\tan a - \tan a\,(1 - \tan^2 a)}{1 - \tan^2 a} = \frac{\tan a + \tan^3 a}{1 - \tan^2 a}.$$

Therefore, each summand can be written as

$$\frac{\tan\bigl(\frac{x}{2^{r+1}}\bigr) + \tan^3\bigl(\frac{x}{2^{r+1}}\bigr)}{1 - \tan^2\bigl(\frac{x}{2^{r+1}}\bigr)} = \tan\!\bigl(\tfrac{x}{2^r}\bigr) \;-\;\tan\!\bigl(\tfrac{x}{2^{r+1}}\bigr).$$

Summing from $$r=0$$ to $$r=n$$ gives a telescoping series:

$$\sum_{r=0}^{n}\Bigl[\tan\!\bigl(\tfrac{x}{2^r}\bigr) - \tan\!\bigl(\tfrac{x}{2^{r+1}}\bigr)\Bigr] = \tan x \;-\;\tan\!\bigl(\tfrac{x}{2^{\,n+1}}\bigr).$$

As $$n \to \infty$$, we have $$\tan\bigl(\tfrac{x}{2^{n+1}}\bigr)\to 0$$. Hence

$$f(x) = \lim_{n \to \infty}\sum_{r=0}^{n}\!\bigl(\cdots\bigr) = \tan x.\quad\text{(Result)}$$

We now evaluate the limit

$$L = \lim_{x \to 0} \frac{e^{x} - e^{f(x)}}{x - f(x)} = \lim_{x \to 0} \frac{e^{x} - e^{\tan x}}{x - \tan x}.$$

As $$x\to 0$$, both numerator and denominator → 0. We use Taylor expansions:

Expansion of tangent:
$$\tan x = x + \frac{x^3}{3} + O(x^5)\quad -(1)$$

Hence
$$x - \tan x = -\frac{x^3}{3} + O(x^5).\quad -(2)$$

Expansion of exponentials:
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4),$$
$$e^{\tan x} = e^{\,x + \frac{x^3}{3} + O(x^5)} = e^x\,e^{\,\frac{x^3}{3} + O(x^5)} = e^x\Bigl(1 + \frac{x^3}{3} + O(x^5)\Bigr).\quad -(3)$$

Subtracting gives
$$e^x - e^{\tan x} = e^x\Bigl[1 - \bigl(1 + \tfrac{x^3}{3} + O(x^5)\bigr)\Bigr] = e^x\Bigl(-\tfrac{x^3}{3} + O(x^5)\Bigr).\quad -(4)$$

Therefore, dividing (4) by (2):

$$\frac{e^x - e^{\tan x}}{x - \tan x} = \frac{e^x\bigl(-\tfrac{x^3}{3} + O(x^5)\bigr)}{-\tfrac{x^3}{3} + O(x^5)} = e^x\;\frac{-\tfrac{x^3}{3} + O(x^5)}{-\tfrac{x^3}{3} + O(x^5)} \;\longrightarrow\; e^0 = 1$$ as $$x \to 0$$.

Final Answer: 1.