The value of $\underset{x\to 1}{lim} \frac{\log x}{\sin \pi x},$ is 

Solution:

We have,

$\begin{array}{l}\underset{x\to 1}{lim} \frac{\log x}{\sin \pi x}=\underset{x\to 1}{lim} \frac{\log [1+(x-1\left)\right\}}{\sin (\pi -\pi x)}\\ =\underset{x\to 1}{lim} \frac{\log \{1+(x-1\left)\right\}}{\sin \pi (1-x)}\\ =\underset{x\to 1}{lim} \frac{\log \{1+(x-1\left)\right\}}{x-1}\times \frac{x-1}{\sin \pi (1-x)}\\ =-\frac{1}{\pi }\underset{x\to 1}{lim} \frac{\log \{1+(x-1\left)\right\}}{x-1}\times -\frac{\pi (1-x)}{\sin \pi (1-x)}=-\frac{1}{\pi }\end{array}$