Question

$\begin{array}{cc}f\left(x\right)=\frac{e^{5x}-e^{2x}}{\sin 3x}& ,forx\ne 0\\ =1,& forx=0\end{array}\}at=0$

examine continuity of the function given in the figure.

Answer 1

$f\left(x\right)=\frac{e^{5x}-e^{2x}}{\sin 3x}$ For $x\ne 0$

$=1$ For $x=0$

$\Rightarrow L.H.L\left(f(0-)\right)={lim}_{x\to 0-}\frac{e^{5x}-e^{2x}}{\sin 3x}$

this is in $\frac{0}{0}$ form

By L' Hospital rule

$={lim}_{x\to 0-}\frac{{5e}^{5x}-{2e}^{2x}}{3\cos 3x}$

$=\frac{5-2}{3}$

$=1$

$\Rightarrow R.H.L\left(f(0+)\right)={lim}_{x\to 0+}\frac{e^{5x}-e^{2x}}{\sin 3x}$

$={lim}_{x\to 0+}\frac{{5e}^{5x}-{2e}^{2x}}{3\cos 3x}$

$=\frac{5-2}{3}$

$=1$

Right hand limit $=$ Left hand limit $=f\left(0\right)$

$\therefore$ The function is continuous at $x=0.$