The function f given by `f(x)={((e^(1//x)-1)/(e^(1//x)+1)",","if",x ne 0),(0",","if", x =0):}`, is

Options

A. discontinuous at x = 0
B. continuous at x = 0
C. continuous everywhere
D. None of the above

Answer

A

Explanation

The left hand limit of f at x = 0 is given by
`lim_(x to0^(-))f(x)=lim_(x to 0^(-))(e^(1//x)-1)/(e^(1//x)+1)=(0-1)/(0+1)=-1`
Similarly, `lim_(x to 0^(+))f(x) =lim_( x to 0^(+))(e^(1//x)-1)/(e^(1//x)+1)`
`=lim_(x to 0^(+))(1-(1)/(e^((1)/(x))))/(1+(1)/(e^((1)/(x)))) =lim_(x to 0^(+)) (1-e^((-1)/(x)))/(1+e^((-1)/(x))) =(1-0)/(1+0)=1`
Thus, `lim_(x to 0^(-))f(x) ne lim_( x to 0^(+))f(x)`, therefore, `lim_(x to 0)f(x)` does not exit. Hence, f is discontinuous at x = 0.