I've been attempting to prove this limit (again). It must be a proof with epsilon delta definition.
$\frac{\sin(\log(x))}{\log(x)} \rightarrow 1$ when $x \rightarrow 1$
so far I have, given $\epsilon >0 $
$\lvert x-1\rvert < \delta$
$\Big \rvert \dfrac{\sin(\log(x))}{\log(x)} -1 \Big \lvert$ < $\Big \rvert \dfrac{\lvert \log(x)\rvert}{\log(x)} -1 \Big \lvert$ =$ \lvert 0 \rvert$
I've used $\lvert \sin(x) \rvert < \lvert x \rvert, x>0$
But this just seems wrong and i would like some help. Thanks in advance.