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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{f}(0) &=\lim _{\mathrm{x} ightarrow 0} \frac{\left(\mathrm{e}^{2 \mathrm{x}}-1ight) \sin \frac{x \pi}{180}}{\mathrm{x}^{2}} \quad\quad\cdots\left[\because x^{\circ}=\left(\frac{x \pi}{180}ight)^{c}ight] \ &=\lim _{\mathrm{x} ightarrow 0}\left(\frac{\mathrm{e}^{2 \mathrm{x}}-1}{\mathrm{x}}ight)\left(\frac{\sin \frac{\pi \mathrm{x}}{180}}{\mathrm{x}}ight) \ &=\left(2 \lim _{\mathrm{x} ightarrow 0} \frac{\mathrm{e}^{2 \mathrm{x}}-1}{2 \mathrm{x}}ight) \frac{\pi}{180}\left[\lim _{\mathrm{x} ightarrow 0} \frac{\frac{\sin \mathrm{x} \pi}{180}}{\frac{\mathrm{x} \pi}{180}}ight]=2(\log \mathrm{e}) 	imes\left(\frac{\pi}{180}ight)=\frac{\pi}{90} \end{aligned}$

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

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