Question

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

Answer 1

We have,

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

So,

$LHL,$

${lim}_{x\to 1-}f\left(x\right)={lim}_{h\to 0}f(1-h)={lim}_{h\to 0}\frac{e^{5(1-h)}-e^{2(1-h)}}{\sin 3(1-h)}$

$={lim}_{h\to 0}\frac{\frac{e^5}{e^{5h}}-\frac{e^2}{e^{2h}}}{\sin (3-3h)}$

Taking limit and we get,

$=\frac{e^5-e^2}{\sin 3}(\because 5h\to 0,2h\to 0,3h\to 0\text{as}h\to 0)$

Now,

$R.H.L$

${lim}_{x\to 1+}f\left(x\right)={lim}_{h\to 0}f(1+h)={lim}_{h\to 0}f(1+h)$

$={lim}_{h\to 0}\frac{e^{5(1+h)}-e^{2(1+h)}}{\sin 3(1+h)}$

$={lim}_{h\to 0}\frac{e^5.e^{5h}-e^2.e^{2h}}{\sin (3+3h)}$

Taking limit and we get,

$=\frac{e^5-e^2}{\sin 3}(\because 5h\to 0,2h\to 0,3h\to 0\text{as}h\to 0)$

Now,

$f\left(1\right)=\frac{e^{5\left(1\right)}-e^{2\left(1\right)}}{\sin 3\left(1\right)}$

$f\left(1\right)=\frac{e^5-e^2}{\sin 3}$

So,

$L.H.L=R.H.L=f\left(x\right)$

Hence, the function is continouos at $x=1$

Hence proved.