Q. 22 - Calculus

Question

Use a geometric argument and the Squeeze Theorem for limits to argue that $\underset{\theta \to 0^{-}}{lim} \sin \theta =0$

for sufficiently small negative angles θ.

Step-by-Step Solution

Answer

The given statement is proved.

Step 1. Given information.

We have to use geometric argument and the Squeeze Theorem for limits to argue $\underset{\theta \to 0^{-}}{lim} \sin \theta =0$

Step 2. Prove the given statement

Look at the picture below:

Using Squeeze Theorem to calculate limits of $\sin \theta$ at x = 0. take angle measured in radians $-\frac{\pi }{4}<\theta <0$

According to the figure, we clearly have $0<\sin \theta <\theta$. Since $\underset{\theta \to 0^{-}}{lim} 0=0$$\underset{\theta \to 0^{-}}{lim} \theta =0$,by Squeeze Theorem we must also have $\underset{\theta \to 0^{-}}{lim} \sin \theta =0$