If $f(x)=\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}} \quad$ if $x \neq 0$ $=k \quad$ if…

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 $k=$
  1. $e$
  2. $\sqrt{e}$
  3. $e^{2}$
  4. $e^{4}$

Solution

Given $f(x)$ is continuous at $x=0$ $\therefore \lim _{x \rightarrow 0}\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}}=K$ $\therefore \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}=K$ $\therefore \frac{\left[\lim _{x \rightarrow 0}(1+\tan x)^{\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}{\left[\lim _{x \rightarrow 0}(1-\tan x) \frac{-1}{\tan x}\right]^{\frac{\tan x}{x}}}=\mathrm{K}$ $\frac{\left[\lim _{x \rightarrow 0}(1+\tan x) \frac{1}{\tan x}\right]^{\lim _{x \rightarrow 0} \frac{\tan x}{x}}}{\left[\lim _{x \rightarrow 0}(1-\tan x)\right]^{-\lim _{x \rightarrow 0} \frac{\tan x}{x}}}=K$ $\frac{e^{1}}{e^{-1}}=K \quad \Rightarrow K=e^{2}$

Asked in: MHT CET 2020 (13 Oct Shift 2)

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