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

Options

A. e
B. `e^(-1)`
C. `e^(2)`
D. `e^(-2)`

Answer

C

Explanation

We have `f(x)=[tan((pi)/(4)+x)]^(1//x)=K`
Since, f(x) is continuous at x=0, then
`f(0)=underset(xto 0)(lim)f(x)`
`=underset(xto0)(lim)[tan((pi)/(4)+x)]^(1//x)`
`impliesK=underset(xto0)(lim)[(1+tanx)/(1-tanx)]^(1//x)` . . . [`1^(oo)` form]
`=e^(underset(xto0)(lim))[(1+tanx)/(1-tanx)-1]*(1)/(x)`
`=e^(underset(xto0)(lim))((2tanx)/(1-tanx))*(1)/(x)`
`=e^(underset(xto0)(2lim))(tanx)/(x)*underset(xto0)(lim)(1)/(1-tanx)`
`k=e^(2.1((1)/(1-0)))=e^(2)" "[becauseunderset(xto0)(lim)(tanx)/(x)=1]`