Let f ( x ) = lim n r = 0 n ( tan ( x / 2 r + 1 ) + tan 3 ( x / 2 r + 1 ) 1 – tan 2 ( x / 2 r + 1 ) ) . Then lim x 0 e x – e f ( x ) ( x – f ( x ) ) is equal to __________. [2025]

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Solution

Q.
Let $f (x) = \lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{\tan (x / 2^{r + 1}) + \tan^{3} (x / 2^{r + 1})}{1 - \tan^{2} (x / 2^{r + 1})})$ . Then $\lim_{x \to 0} \frac{e^{x} - e^{f (x)}}{(x - f (x))}$ is equal to __________. [2025]

Ans.
(1)

We have,

$= \lim_{n \to \infty} \sum_{r = 0}^{n} [\frac{2 \tan (\frac{x}{2^{r + 1}}) - \tan (\frac{x}{2^{r + 1}}) + \tan^{3} (\frac{x}{2^{r + 1}})}{1 - \tan^{2} (\frac{x}{2^{r + 1}})}]$

$= \lim_{n \to \infty} \sum_{r = 0}^{n} [\frac{2 \tan (\frac{x}{2^{r + 1}})}{1 - \tan^{2} (\frac{x}{2^{r + 1}})} - \frac{\tan (\frac{x}{2^{r + 1}}) {1 - \tan^{2} (\frac{x}{2^{r + 1}})}}{{1 - \tan^{2} (\frac{x}{2^{r + 1}})}}]$

$= \lim_{n \to \infty} \sum_{r = 0}^{n} (\tan \frac{x}{2^{r}} - \tan (\frac{x}{2^{r + 1}})) = \tan x$

Now, $\lim_{x \to 0} (\frac{e^{x} - e^{\tan x}}{x - \tan x})$

$= \lim_{x \to 0} e^{\tan x} \frac{(e^{x - \tan x} - 1)}{(x - \tan x)} = 1$ $[∵ \lim_{y \to 0} \frac{e^{y} - 1}{y} = 1]$

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