Question

Examine the continuity of the following function at given points:
(i) $f\left(x\right)=\frac{e^{5x}-e^{2x}}{\sin 3x},forx\ne 0=1,forx\ne 0$

Answer 1

We have,

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

For continuity

L.H.L.

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

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

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

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

${lim}_{x\to 0-}f\left(x\right)=0$

R.H.L.

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

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

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

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

${lim}_{x\to 0-}f\left(x\right)=0$

L.H.L=R.H.L

It is continuous

Hence, this is the answer.