Question

Show that $f\left(x\right)=\left\{\begin{array}{c}\frac{\sin 3x}{\tan 2x},\mathrm{if}x<0\\ \frac{3}{2},\mathrm{if}x=0\\ \frac{\log (1+3x)}{e^{2x}-1},\mathrm{if}x>0\end{array}\right.\mathrm{is}\mathrm{continuous}\mathrm{at}x=0$

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{c}\frac{\sin 3x}{\tan 2x},\mathrm{if}\mathrm{x}<0\\ \frac{3}{2},\mathrm{if}\mathrm{x}=0\\ \frac{\log \left(1+3x\right)}{e^{2x}-1},\mathrm{if}\mathrm{x}>0\end{array}\right.$

We observe
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)$
$$\begin{gathered} =\underset{h\to 0}{\lim }\left(\frac{\sin 3\left(-h\right)}{\tan 2\left(-h\right)}\right)=\underset{h\to 0}{\lim }\left(\frac{\sin 3h}{\tan 2h}\right)=\underset{h\to 0}{\lim }\left(\frac{{\displaystyle \frac{3\sin 3h}{3h}}}{{\displaystyle \frac{2\tan 2h}{2h}}}\right) \\ =\frac{\underset{h\to 0}{\lim }\left({\displaystyle \frac{3\sin 3h}{3h}}\right)}{\underset{h\to 0}{\lim }\left({\displaystyle \frac{2\tan 2h}{2h}}\right)}=\frac{3\underset{h\to 0}{\lim }\left({\displaystyle \frac{\sin 3h}{3h}}\right)}{2\underset{h\to 0}{\lim }\left({\displaystyle \frac{\tan 2h}{2h}}\right)}=\frac{3\times 1}{2\times 1}=\frac{3}{2} \end{gathered}$$


(RHL at x = 1) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }f\left(h\right)$
$$\begin{gathered} =\underset{h\to 0}{\lim }\left(\frac{\log \left(1+3h\right)}{e^{2h}-1}\right)=\underset{h\to 0}{\lim }\left(\frac{3h{\displaystyle \frac{\log \left(1+3h\right)}{3h}}}{{\displaystyle \frac{2h\left(e^{2h}-1\right)}{2h}}}\right) \\ =\frac{3}{2}\underset{h\to 0}{\lim }\left(\frac{{\displaystyle \frac{\log \left(1+3h\right)}{3h}}}{{\displaystyle \frac{\left(e^{2h}-1\right)}{2h}}}\right)=\frac{3}{2}\frac{\underset{h\to 0}{\lim }\left({\displaystyle \frac{\log \left(1+3h\right)}{3h}}\right)}{\underset{h\to 0}{\lim }\left({\displaystyle \frac{\left(e^{2h}-1\right)}{2h}}\right)}=\frac{3\times 1}{2\times 1}=\frac{3}{2} \end{gathered}$$

And, $f\left(0\right)=\frac{3}{2}$


∴ ​$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

Thus, f(x) is continuous at x = 0.