The function `f(x)=x*e^{-(1/|x|+1/x)}` if `x!=0` and `f(x)=0` if `x=0` then

Prove that the function f(x)={x/(1+e^(1/x)); if x != 0 0; if x=0 is not differentiable at x = 0.

On the interval I = [-2, 2], the function f(x)={((x+1)e^(-(1/|x|+1/x)), x ne 0),(0, x=0):}

The function f(x)={(e^(1/x)-1)/(e^(1/x)+1),x!=0, at x=0 ,f(x)=0 a. is continuous at x=0 b. is not continuous at x=0 c. is not continuous at x=0, but can be made continuous at x=0 (d) none of these

The function f(x)=(1+x)^(5//x), f(0)=e^(5)" at x =0 is "

The function f(x)= xcos(1/x^2) for x != 0, f(0) =1 then at x = 0, f(x) is

For the function f (x) = {{:(x/(1+e^(1//x))", " x ne 0),(" 0 , "x = 0):} , the derivative from the right, f'(0^(+)) = … and the derivative from the left, f'(0^(-)) = … .

The function f(x) =(e^(1//x^(2))/(e^(1//x^(2))-1)" for "x ne 0,f(0)=1" at x =0 is"

The function f(x)={(e^(1/x)-1)/(e^(1/x)+1),x!=0 \ \ \ \ \ \ \ 0,x=0 at x=0