Question

Find the value of $k$, if the function $f$ is given by $f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{3}-\tan x}{\pi -3x},& x\ne \frac{\pi }{3}\\ k,& x=\frac{\pi }{3}\end{array}$ is continuous at $x=\frac{\pi }{3}$.

Answer 1

Given the function $f\left(x\right)$ is continuous at $x=\frac{\pi }{3}$.

Then $\underset{x\to \frac{\pi }{3}}{lim}\frac{\sqrt{3}-\tan x}{\pi -3x}=f\left(\frac{\pi }{3}\right)$

Now using L'Hospital's rule we get,

or, $\underset{x\to \frac{\pi }{3}}{lim}\frac{-{\sec }^{2}x}{-3}=f\left(\frac{\pi }{3}\right)$

or, $\frac{4}{3}=f\left(\frac{\pi }{3}\right)$

or, $k=\frac{4}{3}$.