Question

the value of $x\stackrel{lim}{\to }0\frac{\sin x^0}{x}$

Answer 1

We have

${lim}_{x\to 0}\frac{\sin x}{x}=1$ when $x$ is in radians

also, $x^{\circ }=\frac{\pi }{180}x$ radians

$\therefore {lim}_{x\to 0}\frac{\sin x^{\circ }}{x}={lim}_{x\to 0}\frac{\sin \frac{\pi x}{180}}{x}={lim}_{x\to 0}\frac{\sin \left(\frac{\pi x}{180}\right)}{\left(\frac{\pi x}{180}\right)\times \left(\frac{180}{\pi }\right)}\Rightarrow {lim}_{x\to 0}\frac{\sin x^{\circ }}{x}=\frac{\pi }{180}{lim}_{x\to 0}\frac{\sin \left(\frac{\pi x}{180}\right)}{\left(\frac{\pi x}{180}\right)}=\frac{\pi }{180}$