Q. 94 - Calculus

Question

Prove that $\underset{\theta \to 0}{\lim }\frac{1-\cos \theta }{\theta }=0$ by using the double-angle identity $\cos 2\theta =1-2{\sin }^2\theta$ and the other special trigonometric limit $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$.

Step-by-Step Solution

Answer

It is proved that $\underset{\theta \to 0}{\lim }\frac{1-\cos \theta }{\theta }=0$.

Step 1. Given Information

We are given,

 $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$

Step 2. Proving the statement

The limit is given by,

$$\begin{gathered} \underset{\theta \to 0}{\lim }\frac{1-\cos \theta }{\theta }=\underset{\theta \to 0}{\lim }\frac{1-\cos \left(2\frac{\theta }{2}\right)}{\theta } \\ =\underset{\theta \to 0}{\lim }\frac{1-\left[1-2{\sin }^2\left(\frac{\theta }{2}\right)\right]}{\theta } \\ =\underset{\theta \to 0}{\lim }\frac{1-1+2{\sin }^2\left(\frac{\theta }{2}\right)}{\theta } \\ =\underset{\theta \to 0}{\lim }\frac{2{\sin }^2\left(\frac{\theta }{2}\right)}{\theta } \\ =\underset{\theta \to 0}{\lim }\frac{{\sin }^2\left(\frac{\theta }{2}\right)}{\frac{\theta }{2}} \\ =\underset{\theta \to 0}{\lim }\left[\sin \left(\frac{\theta }{2}\right)\frac{\sin \left(\frac{\theta }{2}\right)}{\frac{\theta }{2}}\right] \\ =\sin \left(\frac{0}{2}\right)\left(1\right) \\ =0 \end{gathered}$$

Hence Proved.