Question

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

Answer 1

Consider the given function

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

Then, we have

For R.H.L

${lim}_{x\to \frac{\pi }{6}-}f\left(x\right)={lim}_{x\to \frac{\pi }{6}-}\frac{1-\sqrt{3}\tan x}{\pi -6x}$

$=\frac{1-\sqrt{3}\tan \frac{\pi }{6}}{\pi -6\times \frac{\pi }{6}}=0$

For L.H.L

${lim}_{x\to \frac{\pi }{6}+}f\left(x\right)={lim}_{x\to \frac{\pi }{6}+}\frac{1-\sqrt{3}\tan x}{\pi -6x}$

$=\frac{1-\sqrt{3}\tan \frac{\pi }{6}}{\pi -6\times \frac{\pi }{6}}=0$

And

At $\frac{\pi }{6}$

${lim}_{x\to \frac{\pi }{6}}f\left(x\right)={lim}_{x\to \frac{\pi }{6}}\frac{1-\sqrt{3}\tan x}{\pi -6x}$

$=\frac{1-\sqrt{3}\tan \frac{\pi }{6}}{\pi -6\times \frac{\pi }{6}}=0$

Then,

${lim}_{x\to \frac{\pi }{6}-}f\left(x\right)={lim}_{x\to \frac{\pi }{6}+}f\left(x\right)={lim}_{x\to \frac{\pi }{6}}f\left(x\right)$

Hence, this function is continuous at $x=\frac{\pi }{6}$