Question

${lim}_{x\to 1}\frac{1-\frac{1}{x}}{sin\pi (x-1)}$

Answer 1

${lim}_{x\to 1}\frac{1-\frac{1}{x}}{sin\pi (x-1)}={lim}_{x-1\to 0}\frac{(x-1)}{x\times sin\pi (x-1)}$

As $x\to 1$, then $x-1\to 0$, let $x-1=y\Rightarrow y\to 0$

${lim}_{y\to 0}\frac{y}{(y+1)sin\left(\pi y\right)}$

$={lim}_{y\to 0}\frac{y}{(y+1)\left(sin\pi y\right)}={lim}_{y\to 0}\frac{(y+1)sin\pi y}{y}=\frac{1}{\left({lim}_{y\to 0}\right(y+1\left)\right)\times \left({lim}_{y\to 0}\frac{sin\pi y}{y\times \pi }\right)\times \pi }$

$=\frac{1}{\left(1\right)(1\times \pi )}$ $[\because {lim}_{\theta \to 0}\frac{sin\theta }{\theta }=1]$

$=\frac{1}{\pi }$