Question

Examine the continuity of the function
$f\left(x\right)=\{\begin{array}{cc}\frac{e^{5x}-e^{3x}}{\sin 3x}& \text{for}x\ne 0\\ 1,& \text{for}x=0\end{array}$,
at $x=0$.

Answer 1

$f\left(x\right)=\{\begin{array}{c}\frac{e^{5x}-e^{3x}}{\sin 3x},x\ne 0\\ 1,x=0\end{array}$

Checking continuity at $x=0$

$\Rightarrow \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{e^{5(0+h)}-e^{3(0+h)}}{\sin 3(0+h)}=\underset{h\to 0}{lim}\frac{5e^{5h}-3e^{\left(3h\right)}}{3\cos 3\left(h\right)}=\frac{2}{3}\to 1$ (By L' Hospital Rule)

$\Rightarrow \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{e^{5(0-h)}-e^{3(0-h)}}{\sin 3(0-h)}=\underset{h\to 0}{lim}\frac{-5e^{-5h}+3e^{-3h}}{3\cos 3(-h)}=\frac{-2}{3}\to 2$ (By L' Hospital Rule)

$\Rightarrow f\left(0\right)=1\to 3$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$

$\underset{x\to 0^{+}}{lim}f\left(x\right)\ne \underset{x\to 0^{-}}{lim}f\left(x\right)\ne f\left(1\right)$

$\Rightarrow f\left(x\right)$ is discontinuous at $x=1$