Question

Examine the continuity of the following functions at given points
$f\left(x\right)=\frac{e^{5x}-e^{2x}}{sin3x}$, for $x\ne 0$
$=1$ for $x=0$ $\}$ at $x=0$

Answer 1

$LHL={lim}_{h\to 0}f(0-h){lim}_{h\to 0}\left(\frac{e^{-5h}-e^{-2h}}{sin(-3h)}\right){lim}_{h\to 0}\left(\frac{e^{-2h}(e^{-3h}-1)}{-sin\left(3h\right)}\right)={lim}_{h\to 0}{e}^{-2h}\left(\frac{(e^{-3h}-1)}{-\left(3h\right)}\right)\times \left(\frac{-3h}{-sin3h}\right)=e^0\times 1\times 1=1RHL{lim}_{h\to 0}f(0+h)={lim}_{h\to 0}\left(\frac{e^{5h}-e^{2h}}{sin\left(3h\right)}\right)00000={lim}_{h\to 0}{e}^{2h}\left(\frac{(e^{3h}-1)}{\left(3h\right)}\right)\times \left(\frac{3h}{sin3h}\right)=e^0\times 1\times 1=1asLHL=RHL=f\left(0\right),hencef\left(x\right)iscontinuousatx=0$