Question

Prove that $\underset{x\to 0}{lim}\left[\frac{\sin x}{x}\right]=1$ where $x$ is in radians and hence evaluate $\underset{x\to 0}{lim}\left[\frac{\sin ax}{bx}\right]$.

Answer 1

To prove: $\underset{x⟶0}{lim}\frac{\sin x}{x}=1$

Put $=0$ then

$\frac{\sin 0}{0}=\frac{0}{0}$

In this form we can approch to L-Hospital rule,

Now differentiate both numerator and the denominator with respect to $x$.

$\underset{x⟶0}{lim}\frac{\frac{d}{dx}\left(\sin x\right)}{\frac{d}{dx}\left(x\right)}=\underset{x⟶0}{lim}\frac{\cos x}{1}$

Now, put $x=0$

$\underset{x⟶0}{lim}\frac{\cos x}{1}=\frac{\cos 0}{1}=1$

Hence, $\underset{x⟶0}{lim}\frac{\sin x}{x}=1$ proved.

if $x\to 0$ implies $ax\to 0$, we have

$\underset{x\to 0}{lim}\frac{\sin ax}{bx}$

$=\underset{ax\to 0}{lim}\left(\frac{a}{b}\right)\frac{\sin ax}{ax}$

$=\frac{a}{b}\times \underset{ax\to 0}{lim}\frac{\sin ax}{ax}$

$=\frac{a}{b}\times 1$

$=\frac{a}{b}$