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 \(k=\)

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