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Given:

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

We observe

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$

$=\lim _{h \rightarrow 0}\left(\frac{\sin 3(-h)}{\tan 2(-h)}\right)=\lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{\tan 2 h}\right)=\lim _{h \rightarrow 0}\left(\frac{\frac{3 \sin 2 h}{3 h}}{\frac{2 \tan 2 h}{2 h}}\right)$

$=\frac{\lim _{h \rightarrow 0}\left(\frac{3 \sin 3 h}{3 h}\right)}{\lim _{h \rightarrow 0}\left(\frac{2 \tan 2 h}{2 h}\right)}=\frac{3 \lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{3 h}\right)}{2 \lim _{h \rightarrow 0}\left(\frac{\tan 2 h}{2 h}\right)}=\frac{3 \times 1}{2 \times 1}=\frac{3}{2}$

$(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)$

$=\lim _{h \rightarrow 0}\left(\frac{\log (1+3 h)}{e^{2 h}-1}\right)=\lim _{h \rightarrow 0}\left(\frac{3 h \frac{\log (1+3 h)}{3 h}}{\frac{2 h\left(e^{2 h-1}\right)}{2 h}}\right)$

$=\frac{3}{2} \lim _{h \rightarrow 0}\left(\frac{\frac{\log (1+3 h)}{3 h}}{\frac{\left(e^{2 h-1}\right)}{2 h}}\right)=\frac{3}{2} \frac{\lim _{h \rightarrow 0}\left(\frac{\log (1+3 h)}{3 h}\right)}{\lim _{h \rightarrow 0}\left(\frac{\left(e^{2 h_{-1}}\right)}{2 h}\right)}=\frac{3 \times 1}{2 \times 1}=\frac{3}{2}$

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

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

Thus, f(x) is continuous at x = 0.