Prove that a function is differentiable at a point At which values of $x$ is $f(x)$ differentiable? 
$f(x) =
\begin{cases}
1-e^{-x},  & \text{$x \gt 0$} \\
\ln(1-x), & \text{$x\le 0$}
\end{cases}$
I first proved that $f(x)$ is continuous for every $x$:
$$
\lim_{x \to 0^+}f(x) = 1-e^{-0} = 0
$$
$$
\lim_{x \to 0^-}f(x) = \ln(1-x) = 0
$$
and $f(0) = 0$, is this OK? was that necessary when proving that a function is differentiable at a point? 
Next, I wanna prove the limit using the definition of derivative, so:
$$
\lim_{x\to0^+} \frac{f(x)-f(0)}{x-0} = \frac{1-e^{-x}-(1-e^0)}{x} = \frac{-e^{-x}+e^0}{x}
$$ 
$$
\lim_{x\to0^-} \frac{f(x)-f(0)}{x-0} = \frac{\ln(1-x)-0}{x} = \frac{\ln(1-x)}{x}
$$ 
How do I continue from here? I'm stuck at both sides, some help please?:)
Thanks!
 A: Your work is correct. Notice that the limit on the right and on the left are the derivative of $1-e^{-x}$ and $\ln(1-x)$ respectively at $0$ so you need to verify that they are equal.
A: Note that you have an error: by definition, $f(0)=\ln(1-0)=0$, and that happens on both "sides" of zero. So, 
$$
\lim_{x\to0^+} \frac{f(x)-f(0)}{x-0} = \frac{1-e^{-x}-\color{red}{0}}{x} = \frac{1-e^{-x}-\color{red}{0}}{x}\qquad(1)
$$ 
$$
\lim_{x\to0^-} \frac{f(x)-f(0)}{x-0} = \frac{\ln(1-x)-0}{x} = \frac{\ln(1-x)}{x}\qquad(2)
$$ 
In the first limit, note the numerator tends to $0$ and the denominator to $0$, so we can use L'Hôpital's Rule:
$$ (1)\qquad\lim_{x\to0^+}\frac{1-e^{-x}}{x}\overbrace{=}^{\text{LHR}}\lim_{x\to0^+}\frac{e^{-x}}{1}=1$$ 
For the second limit, note that is of the form $\frac{0}{0}$, so we use L'Hôpital's Rule:
$$ (2)\qquad \lim_{x\to0^-}\frac{\ln(1-x)}{x}\overbrace{=}^{\text{LHR}}\lim_{x\to0^-}\frac{-\frac{1}{1-x}}{1}=\lim_{x\to0^-}\frac{1}{x-1}=\frac{1}{0-1}=-1$$ 
Both limits are different, so $f(x)$ is not differentiable at $x=0$.
A: Using the fact that, $\lim_{x\to0}\dfrac{a^x-1}{x}=\log_ea$ (check out when $(a = e)$!)  
Limit-1:  
$\lim_{x\to0^+}\dfrac{-e^{-x}+e^0}{x}=\lim_{x\to0^+}\dfrac{1-e^{-x}}{x}=\lim_{x\to0^+}\dfrac{e^{-x}-1}{-x}=1$  
Limit-2 
$\lim_{x\to0^-}\dfrac{\ln(1-x)}{x}$ , (put $ln(1-x)=t$ so that when $x\to0^-$ gives $t\to 0^+$)
Simplifying expression in terms of $t$  
$\lim_{t\to0^+}\dfrac{t}{1-e^t}=\lim_{t\to0^+}\dfrac{1}{\dfrac{1-e^t}{t}}=\dfrac{-1}{\lim_{t\to0^+}\left(\dfrac{e^t-1}{t}\right)}=-1$
