Find $\lim \limits_{x \to 1^{+} }\left(\frac{82\sin(\pi x)}{1-x}\right)$ without L'hopital

Question

Is exist the way to calculate:

$$ \lim_{x \to 1^{+} }\frac{82sin(\pi x)}{1-x} $$

Without using the L'hopital rule. From L'hopital rule i got $82\pi$

Thanks a lot!

Answer 1

as $\sin(\pi) = 0$ the limit is the derivative of $x\to \sin(\pi x)$ in $x=1$: $$ -82 \lim_{x\to 1} \frac{\sin (\pi x) - \sin(\pi)}{x - 1} = -82\pi\cos(\pi) = 82\pi $$

Answer 2

You can also solve the limit by a couple of substitutions. Letting $y=1-x$ we have $$\lim_{y\to 0^-}\frac{82\sin(\pi-\pi y)}{y}=\lim_{y\to 0^-}\frac{82(\sin\pi\cos(\pi y)-\sin(\pi y)\cos\pi)}{y}=\\ \lim_{y\to0^-} \frac{-82\sin(\pi y)}{y}=82\cdot\lim_{y\to 0^+}\frac{\sin(\pi y)}{y}.$$ Let $z=\pi y$ to finally obtain $$82\lim_{z\to 0^+}\frac{\sin z}{z/\pi}=82\pi\lim_{z\to 0^+}\frac{\sin z}{z}=82\pi. $$