If $\displaystyle f(x)=\left(\frac{1-x}{1+x}\right)^{\frac{1}{x}}$;$\displaystyle x\neq 0$
and $\displaystyle f(0)=\frac{1}{e^2}$
Evaluate $\displaystyle f_{+}'(0)$ and $\displaystyle f_{-}'(0)$
If $\displaystyle f(x)=\left(\frac{1-x}{1+x}\right)^{\frac{1}{x}}$;$\displaystyle x\neq 0$
and $\displaystyle f(0)=\frac{1}{e^2}$
Evaluate $\displaystyle f_{+}'(0)$ and $\displaystyle f_{-}'(0)$
The standard definition of a one-sided derivative of f at 0 requires f to have a value at 0, which your f does not. What definition of one-sided derivative are you using? Do you mean the value that the derivative function approaches at 0 (ie., $\displaystyle f_{+}'(0) = \lim_{x\rightarrow 0^+} f'(x)$)?