Question

If $f\left(x\right)=\frac{1-\sqrt{3}\tan x}{\pi -6x}$, for $x\ne \frac{\pi }{6}$ is continuous at $x=\frac{\pi }{6}$, find $f\left(\frac{\pi }{6}\right)$.

Answer 1

Given, $f\left(x\right)=\frac{1-\sqrt{3}\tan x}{\pi -6x}$, for $x\ne \frac{\pi }{6}$.

Since the function $f\left(x\right)$ is given to be continuous at $x\ne \frac{\pi }{6}$.

Then

$\underset{x\to \frac{\pi }{6}}{lim}f\left(x\right)=f\left(\frac{\pi }{6}\right)$........(1).

Now,

$\underset{x\to \frac{\pi }{6}}{lim}f\left(x\right)$

$=\underset{x\to \frac{\pi }{6}}{lim}\frac{1-\sqrt{3}\tan x}{\pi -6x}$ $\left(\frac{0}{0}\right)$ form.

Now applying L'Hospital's rule we get,

$=\underset{x\to \frac{\pi }{6}}{lim}\frac{-\sqrt{3}{\sec }^{2}x}{-6}$

$=\frac{2}{3\sqrt{3}}$.

From (1) we have $f\left(\frac{\pi }{6}\right)=\frac{2}{3\sqrt{3}}$.