Question ID: #498
Let $f(x)=\lim_{n\rightarrow\infty}\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1})+\tan^{3}(x/2^{r+1})}{1-\tan^{2}(x/2^{r+1})}\right)$. Then $\lim_{x\rightarrow0}\frac{e^{x}-e^{f(x)}}{(x-f(x))}$ is equal to:
Solution:
$f(x) = \lim_{n\to\infty} \sum (\tan \frac{x}{2^r} – \tan \frac{x}{2^{r+1}})$.
If this is the telescoping form, then the sum is $\tan(x/2^0) – \lim_{n\to\infty} \tan(x/2^{n+1}) = \tan x – 0 = \tan x$.
So, $f(x) = \tan x$.
Now evaluate the limit:
$$L = \lim_{x\to0} \frac{e^x – e^{\tan x}}{x – \tan x}$$
$$L = \lim_{x\to0} e^{\tan x} \left[ \frac{e^{x-\tan x} – 1}{x – \tan x} \right]$$
As $x \to 0$, $x – \tan x \to 0$.
Using the standard limit $\lim_{t\to0} \frac{e^t – 1}{t} = 1$:
$$L = e^0 \cdot 1 = 1$$
Ans. (1)
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