1. ## Limit Help

How do you solve these limits?

$\displaystyle \lim_{x\to0} \ln\frac{e^x-1}{x}$

and

$\displaystyle \lim_{x\to0} \frac{\sin(3x)}{5x}$

2. Originally Posted by Mike9182
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

$\displaystyle \lim_{x\to 0}\frac{\sin(3x)}{5x}=\lim_{x\to 0}\frac{\cos(3x)\cdot 3}{5}=\frac{\cos(3\cdot 0)\cdot 3}{5}=\frac{3}{5}$

3. ok, I must be supposed to graph them then because they showed up in chapter 2 of my book which is the same chapter limits were introduced in.