The value of limx→1 log⁡xsin⁡πx, is - Infinity Learn by Sri Chaitanya

A
$\frac{1}{π}$
B
$- π$
C
$π$
D
$- \frac{1}{π}$

Solution:

We have,

$\begin{array}{l} lim_{x \to 1} \frac{\log x}{\sin π x} = lim_{x \to 1} \frac{\log [1 + (x - 1)}}{\sin (π - π x)} \\ = lim_{x \to 1} \frac{\log {1 + (x - 1)}}{\sin π (1 - x)} \\ = lim_{x \to 1} \frac{\log {1 + (x - 1)}}{x - 1} \times \frac{x - 1}{\sin π (1 - x)} \\ = - \frac{1}{π} lim_{x \to 1} \frac{\log {1 + (x - 1)}}{x - 1} \times - \frac{π (1 - x)}{\sin π (1 - x)} = - \frac{1}{π} \end{array}$

The value of $lim_{x \to 1} \frac{\log x}{\sin π x},$ is

Solution:

Related content