Question

Evaluate the limit : $\underset{x\to 0}{lim}\frac{\sin \left(m{x}^0\right)}{\sin \left(n{x}^0\right)}$

Answer 1

(say) $l=\underset{x\to 0}{lim}\frac{\sin \left(mx°\right)}{\sin \left(nx°\right)}$

$=\underset{x\to 0}{lim}\frac{\frac{\sin \left(mx°\right)}{mx°}.mx°}{\frac{\sin \left(nx°\right)}{nx°}.nx°}$

$=\underset{x\to 0}{lim}\frac{\frac{\sin \left(mx°\right)}{mx°}}{\frac{\sin \left(nx°\right)}{nx°}}\left(\frac{m}{n}\right)$

$=\frac{\underset{x\to 0}{lim}\frac{\sin \left(mx°\right)}{mx°}}{\underset{x\to 0}{lim}\frac{\sin \left(nx°\right)}{nx°}}\left(\frac{m}{n}\right)[\because \underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{lim}f\left(x\right)}{\underset{x\to a}{lim}g\left(x\right)}]$

$=\frac{\underset{x\to 0}{lim}\frac{\sin \left(mx°\right)}{mx°}}{\underset{x\to 0}{lim}\frac{\sin \left(nx°\right)}{nx°}}\left(\frac{m}{n}\right)$ [ multiplying $x°$ and x on numerator and denominator]

$=\frac{\frac{\pi }{180}}{\frac{\pi }{180}}\left(\frac{m}{n}\right)\left[\underset{x\to 0}{lim}\frac{\sin x°}{x}=\frac{\pi }{180}\right]$ [due of the $°$ being changed to radian]

$=\frac{m}{n}$