If \(f(x)=\frac{\left(e^{2 x}-1\right) \sin x^{0}}{x^{2}}, x \neq 0\) is continuous at \(x=0\), then \(f(0)=\)

\(\mathrm{f}(0) =\lim _{\mathrm{x} \rightarrow 0} \frac{\left(\mathrm{e}^{2 \mathrm{x}}-1\right) \sin \frac{x \pi}{180}}{\mathrm{x}^{2}} \quad\quad\cdots\left[\because x^{\circ}=\left(\frac{x \pi}{180}\right)^{c}\right] \)
\( =\lim _{\mathrm{x} \rightarrow 0}\left(\frac{\mathrm{e}^{2 \mathrm{x}}-1}{\mathrm{x}}\right)\left(\frac{\sin \frac{\pi \mathrm{x}}{180}}{\mathrm{x}}\right) \)
\( =\left(2 \lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{e}^{2 \mathrm{x}}-1}{2 \mathrm{x}}\right) \frac{\pi}{180}\left[\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{\sin \mathrm{x} \pi}{180}}{\frac{\mathrm{x} \pi}{180}}\right]=2(\log \mathrm{e}) \times\) \(\left(\frac{\pi}{180}\right)=\frac{\pi}{90}\)