If `f(x)=sqrt(1-e^(-x^2))`, then at `x=0f(x)` is

Options

A. differentiable as well as continuous
B. continuous but not differentiable
C. differentiable but not continuous
D. neigther differentiable nor continuous

Answer

B

Explanation

Lf (0)
`=lim_(hrarr0)(f(0-h)-f(0))/(-h)=lim_(hrarr0)(sqrt(1-e^(-h^(2))))/(-h)`
`=lim_(hrarr0)([1-(1-h^(2)+(h^(4))/(2!)-…)]^(1//2))/(-h)`
`=lim_(hrarr0)(h.[1-(h^(4))/(2!)+…]^(1//2))/(-h)=-1`
`Rf^(')(0)=lim_(hrarr0)(f(0+h)-f(0))/h=lim_(hrarr0)(sqrt(1-e^(-h^(2))))/h=1`
Hence `f(x)` is not differentiable at `x=0` Since `Lf^(')(0)=Rf^(')(0)=0` (finite) therefore `f(x)` is continuous at `x=0`.
Hence `f(x)` is continuous but not differentiable at `x=0`