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If $f(x)=\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}} \quad$ if $x \neq 0$
$=k \quad$ if $x=0$,
is continuous at $x=0$ then $k=$
Options:
$=k \quad$ if $x=0$,
is continuous at $x=0$ then $k=$
Solution:
1648 Upvotes
Verified Answer
The correct answer is:
$e^{2}$
Given $f(x)$ is continuous at $x=0$
$\therefore \lim _{x \rightarrow 0}\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}}=K$
$\therefore \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}=K$
$\therefore \frac{\left[\lim _{x \rightarrow 0}(1+\tan x)^{\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}{\left[\lim _{x \rightarrow 0}(1-\tan x) \frac{-1}{\tan x}\right]^{\frac{\tan x}{x}}}=\mathrm{K}$
$\frac{\left[\lim _{x \rightarrow 0}(1+\tan x) \frac{1}{\tan x}\right]^{\lim _{x \rightarrow 0} \frac{\tan x}{x}}}{\left[\lim _{x \rightarrow 0}(1-\tan x)\right]^{-\lim _{x \rightarrow 0} \frac{\tan x}{x}}}=K$
$\frac{e^{1}}{e^{-1}}=K \quad \Rightarrow K=e^{2}$
$\therefore \lim _{x \rightarrow 0}\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}}=K$
$\therefore \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}=K$
$\therefore \frac{\left[\lim _{x \rightarrow 0}(1+\tan x)^{\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}{\left[\lim _{x \rightarrow 0}(1-\tan x) \frac{-1}{\tan x}\right]^{\frac{\tan x}{x}}}=\mathrm{K}$
$\frac{\left[\lim _{x \rightarrow 0}(1+\tan x) \frac{1}{\tan x}\right]^{\lim _{x \rightarrow 0} \frac{\tan x}{x}}}{\left[\lim _{x \rightarrow 0}(1-\tan x)\right]^{-\lim _{x \rightarrow 0} \frac{\tan x}{x}}}=K$
$\frac{e^{1}}{e^{-1}}=K \quad \Rightarrow K=e^{2}$
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