How do you solve these limits?
$\displaystyle \lim_{x\to0} \ln\frac{e^x-1}{x}$
and
$\displaystyle \lim_{x\to0} \frac{\sin(3x)}{5x}$
Usually one would solve them using L'Hospital's Rule like this:
$\displaystyle \lim_{x\rightarrow 0}\frac{\mathrm{e}^x-1}{x}=\lim_{x\to 0}\frac{\mathrm{e}^x}{1}=\frac{\mathrm{e}^0}{1}=1$
EDIT: I have simply dropped the $\displaystyle \ln$ so let me do it (hopefully) right
$\displaystyle \lim_{x\to0} \ln\frac{e^x-1}{x}=\ln \lim_{x\to0} \frac{e^x-1}{x}=\ln \lim_{x\to 0}\frac{\mathrm{e}^x}{1}=\ln\frac{\mathrm{e}^0}{1} =\ln 1=0$
END OF EDIT
and
$\displaystyle \lim_{x\to 0}\frac{\sin(3x)}{5x}=\lim_{x\to 0}\frac{\cos(3x)\cdot 3}{5}=\frac{\cos(3\cdot 0)\cdot 3}{5}=\frac{3}{5}$