Compute this limit by applying this question Show that : $ \frac{f'(x)}{g'(x)}\xrightarrow[x \to a]{} l \implies \frac{f(x)}{g(x)}\xrightarrow[x \to a]{} l\quad (l\in \mathbb{\bar{R}}) $
$$\lim _{x\to \:1+}\left(\frac{\sin \left(\pi x\right)}{\ln \left(x\right)}\right)$$
My thoughts:
$\lim _{x\to \:1+}\left(\frac{\sin \left(\pi x\right)}{\ln \left(x\right)}\right)$
Applying l'hopital rule:
$=\lim _{x\to \:1+}\left(\frac{\cos \left(\pi x\right)\pi }{\frac{1}{x}}\right)$ $=\lim _{x\to \:1+}\left(\pi x\cos \left(\pi x\right)\right)$ $=\pi \lim _{x\to \:1+}\left(x\right)\lim _{x\to \:1+}\left(\cos \left(\pi x\right)\right)$ $=\pi 1\left(-1\right)$ $=-\pi $
but if we want to apply question which i think is weaker version of hopital rule how we gonna do
Maybe I should have written this in a comment, but it is a bit too long. Since the hypothesis of De l'Hôpital's rule (Bernoulli Theorem) are a bit awkward and difficult to remember, many mathematicians prefer to completely avoid it and use Taylor expansions with Peano remainders, instead. In this case, since in a neighbourhood of the origin: $$\sin z=z+o(z^2),\qquad \log(1+z)=z-\frac{z^2}{2}+o(z^2), $$ we have: $$ \lim_{x \to 1}\frac{\sin(\pi x)}{\log(x)}=\lim_{z\to 0}\frac{-\sin(\pi z)}{\log(1+z)}=\lim_{z\to 0}\frac{-\pi z + o(z^2)}{z+O(z^2)}=\color{red}{-\pi}, $$ without explicitly differentiating any function.