Another one:
this time im lost
Recall, L'Hospital's rule states that if
$\displaystyle \displaystyle \lim_{x\to n}\frac{f(x)}{g(x)} = \frac{0}{0} \;\;\text{or}\;\; \frac{\pm \infty}{\pm \infty}\quad \text{then}\quad \lim_{x\to n} \frac{f(x)}{g(x)} \overset{\text{H}}{=}\lim_{x\to n}\frac{f'(x)}{g'(x)}$
So....
$\displaystyle \displaystyle \lim_{x\to 1}\frac{\ln(x)}{\sin(2\pi x)}\overset{\text{H}}{=}\lim_{x\to 1}\frac{\frac{d}{dx}\ln(x)}{\frac{d}{dx}\sin(2\pi x)}$
I'm confident you can handle it from here.