Derivative of limit using L'Hôpital's rule Original question is: $$\lim_{x \to 0}\frac {1}{x} \cdot \ln (e^x+x)$$
Which is $$\frac 00$$
So I use L'Hôpital's rule and use the derivative but I'm not sure how to do that. 
I tried and got: $$-\frac {1}{x^2}\cdot\frac{1}{e^x+x}$$
which doesnt help me at all...
Thanks
 A: l'Hopital's rule says that if $f\to0$ and $g\to0$ (at the same point $a$) and both functions are derivable at this point, then
$$\lim_{x\to a}\frac fg=\lim_{x\to a}\frac{f'}{g'}$$
Your limit can be written this way:
$$\lim_{x\to0}\frac{\ln(e^x+x)}{x}$$
Since numerator and denominator tend to $0$, this is the same as
$$\lim_{x\to0}\frac{\frac{e^x+1}{e^x+x}}{1}=2$$
A: Hint:  using the chain rule,
$$
\frac{d}{dx} \ln ( e^x + x ) = \frac{\frac{d}{dx} (e^x + x)}{e^x + x} = \frac{e^x + 1}{e^x + x}.
$$
A: Using L'Hôpital's rule
$$ \lim\limits_{x\to 0} \frac{\ln(e^x+x)}{x} = \lim\limits_{x\to 0} \frac{\frac{\mathrm d}{\mathrm dx}\left[\ln(e^x+x)\right]}{\frac{\mathrm d}{\mathrm dx}[x]}= \lim\limits_{x\to 0} \frac{\frac{1}{e^x+x}\frac{\mathrm d}{\mathrm dx}\left[e^x+x\right]}{1}= \lim\limits_{x\to 0} \frac{e^x+1}{e^x+x}=\frac{1+1}{1+0}=2 $$
Using equivalent infinitesimals
$$\lim\limits_{x\to 0} \frac{\ln(e^x+x)}{x}=\lim\limits_{x\to 0} \frac{e^x+x-1}{x}=\lim\limits_{x\to 0} \left[1+\frac{e^x-1}{x}\right]=1+\lim\limits_{x\to 0} \frac{x\ln e}{x}=2$$
A: $$\lim_{x \to 0}\frac{\ln (e^x+x)}{x}\underbrace{=}_{\mathrm{L'Hospital's rule}}\lim_{x\to 0} \frac{\displaystyle \frac{d}{dx} \frac{e^x+1}{e^x+x}}{\displaystyle \frac{d}{dx} x}=\lim_{x\to 0} \frac{\displaystyle \frac{e^x+1}{e^x+x}}{1}=2.$$
