is the function f(x) = (3x+4tanx)/x continuous at x = 0? if not how may the function be defined to make it continuous at this point?

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{3\mathrm{x}+4\mathrm{tanx}}{\mathrm{x}} \\ \mathrm{As} \mathrm{denominator} \mathrm{becomes} 0 \mathrm{at} \mathrm{x}=0, \mathrm{hence} \mathrm{function} \mathrm{is} \mathrm{not} \mathrm{continuous} \mathrm{at} \mathrm{x}=0. \\ \mathrm{Hence} \mathrm{make} \mathrm{function} \mathrm{continuous} \mathrm{at} \mathrm{x}=0, \mathrm{we} \mathrm{should} \mathrm{define} \mathrm{f}\left(0\right) \mathrm{such} \mathrm{that}: \\ \underset{\mathrm{x}\to 0}{\lim } \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right) \\ \mathrm{Now} \\ \underset{\mathrm{x}\to 0}{\lim } \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{3\mathrm{x}+4\mathrm{tanx}}{\mathrm{x}}\right) \\ =\underset{\mathrm{x}\to 0}{\lim }\left(\frac{3\mathrm{x}}{\mathrm{x}}+\frac{4\mathrm{tanx}}{\mathrm{x}}\right) \\ =\underset{\mathrm{x}\to 0}{\lim }\left(3+\frac{4\mathrm{tanx}}{\mathrm{x}}\right) \\ =3+4\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{tanx}}{\mathrm{x}} \\ =3+4 \left[\mathrm{As} \underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{tanx}}{\mathrm{x}}=1\right] \\ \Rightarrow \underset{\mathrm{x}\to 0}{\lim } \mathrm{f}\left(\mathrm{x}\right)=7 \\ \Rightarrow \mathrm{f}\left(0\right)=7 \\ \mathrm{Hence} \\ \mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\left\{\begin{array}{ll}\frac{\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{tanx}}{\mathbf{x}}& \mathbf{x}\mathbf{\ne }\mathbf{0}\\ \mathbf{7}& \mathbf{x}\mathbf{=}\mathbf{0}\end{array}\right. \end{gathered}$$

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