Q. 23 - Calculus

Question

Use a geometric argument and the Squeeze Theorem for limits to argue that $\underset{\theta \to 0^{-}}{lim} \cos \theta =1$

for sufficiently small negative angles θ. 

Step-by-Step Solution

Answer

$\underset{\theta \to 0^{-}}{lim} \cos \theta =1$ for sufficiently small negative angles θ has been proven.

Step 1. Given information.

We have to use a  geometric argument and the Squeeze Theorem for limits to argue that $\underset{\theta \to 0^{-}}{lim} \cos \theta =1$

Step 2. Prove the argument

Look at the picture below :

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

According to the figure, we clearly have $0<1-\cos \theta <\theta$

Thus by Squeeze Theorem, we have,

$\begin{array}{r}\underset{\theta \to 0^{-}}{lim} (1-\cos \theta )=\underset{\theta \to 0^{-}}{lim} 1-\underset{\theta \to 0^{-}}{lim} \cos \theta \\ =1-\cos 0\\ =1-1\\ =0\end{array}$

Step 3. Rewrite the limits

Rewrite the limits:

$\begin{array}{r}\underset{\theta \to 0^{-}}{lim} (1-\cos \theta )=0\\ \underset{\theta \to 0^{-}}{lim} 1-\underset{\theta \to 0^{-}}{lim} \cos \theta =0\\ 1-\underset{\theta \to 0^{-}}{lim} \cos \theta =0\\ \underset{\theta \to 0^{-}}{lim} \cos \theta =1\end{array}$