`f(x)=[tan(pi/4+x)]^(1/x), x!=0` and `f(x)=k, x=0` is continuous at x=0 then k=

Explanation

Given, `f(x) = [tan(pi/4+x)]^(1/x) " for " x ne 0`
for x = 0
= k
The function would be continuous if `underset(x to 0) lim f (x) = f (0)`
` underset(x to 0) lim [tan (pi/4 +x)]^(1/x) = k`
` underset(x to 0) lim [(1+tan x)/(1-tan x)]^(1/x) (4) = k`
` underset(x to 0) lim [1 + (1+tan x)/(1-tan x) -1] ^(1/x) = k`
` underset(x to 0) lim [1+(1+tan x -1 + tan x)/(1-tan x)]^(1/x) = k`
` underset(x to 0) lim [1+(2 tan x)/(1 -tan x)]^(1/x) = k`
`underset(x to 0) lim [1 + (2 tan x)/(1 - tan x)]^(1/((2 tan x)/(1-tan x))xx(2 tan x)/(x(1-tan x)))=k`
We know that,
`underset(x to 0) lim [1 + x ] ^(1/x) = e`
` e^(underset(x to 0) lim 2(tan x)/(x(1-x)) )= k`
`e^(underset(2x to 0) lim 2 xx (tan x)/x underset(x to 0) lim 1/((1-x)))=k`
`[:' underset(x to 0)lim ((tan x)/x) = 1]`
` e^(2 xx 1 xx 1) = k`
` :. " " k = e^(2) `