Show that f(x)= 3x 2x ,&if x 0 is discontinuous at x = 0.

Given, f(x)= 3x 2x ,&if x 0 We observe (LHL at x=0)= _ x 0^- f(x) = _ h 0 f(0-h)= _ h 0 f(-h) = _ h 0 ( 3(-h) 2(-h) )= _ h 0 ( 3h 2h ) = _ h 0 ( 3 3h 3h 2 2h 2h ) = _ h 0 ( 3 3h 3h ) _ h 0 ( 2 2h 2h ) = 3 _ h 0 ( 3h 3h ) 2 _ h 0 ( 2h 2h ) =3 1 2 1 =3 2 (RHL at x=0)= _ x 0^+ f(x) = _ h 0 f(0+h)= _ h 0 f(h) = _ h 0 ( (1+3h) e^ 2h -1 ) = _ h 0 ( 3h (1+3h) 3h 2h (e^ 2h-1 ) 2h ) 3 2 _ h 0 ( (1+3h) 3h (e^ 2h -1) 2h ) =3 2 _ h 0 ( (1+3h) 3h ) _ h 0 ( (e^ 2h-1 ) 2h ) =3 1 2 1 =3 2 And, f(0)=3 2 _ h 0 f(x)= _ h 0^+ f(x)=f(0) Thus, f(x) is continuous at x = 0.

Show that f(x)= 3x 2x ,&if x<0 3 2 ,&if x = 0 (1+3x) e^ 2x ,&if x…

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Question

Show that $\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$ is discontinuous at x = 0.

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Answer

Given,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$
We observe
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3(-\text{h})}{\tan2(-\text{h})}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{\tan2\text{h}}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{3\sin3\text{h}}{3\text{h}}}{\frac{2\tan2\text{h}}{2\text{h}}}\Bigg)$
$=\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{3\sin3\text{h}}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{2\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{3\text{h}}\Big)}{2\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{\text{e}^{2\text{h}}-1}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{3\text{h}\frac{\log(1+3\text{h})}{3\text{h}}}{2\text{h}\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}}\Bigg)$
$\frac{3}{2}\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{\log(1+3\text{h})}{3\text{h}}}{\frac{(\text{e}^{2\text{h}}-1)}{2\text{h}}}\Bigg)$
$=\frac{3}{2}\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
And, $\text{f}(0)=\frac{3}{2}$
$\therefore\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0.