Question

Find the value of $k$ which the following function is continous at $x=0$
$f\left(x\right)=\frac{\sin 2x}{5x},x\ne 0$
$k,x=0$

Answer 1

$LHL$

$\underset{x\to 0}{lim}f\left(x\right)$

$\underset{x\to 0}{lim}f(0-h)$

$\underset{x\to 0}{lim}f(-h)$

$\underset{x\to 0}{lim}\frac{\sin 2(-h)}{5(-h)}$

$\underset{x\to 0}{lim}\frac{-sin2h}{-5h}$

$\underset{x\to 0}{lim}\frac{2\sinh \cos h}{5h}$

$=\frac{2}{5}\times 1$

$=\frac{2}{5}$

A function is continuous at $x=0$ So,

$k=\frac{2}{5}$

$RHL$

$\underset{x\to 0^{+}}{lim}f\left(x\right)$

$\underset{x\to 0}{lim}f(0+h)$

$\underset{x\to 0}{lim}f\left(h\right)$

$\underset{x\to 0}{lim}k$

$=k$