This explanation comes from my textbook:
In the computation of $\displaystyle{\lim_{\theta \to 0}} \frac{\sin \theta}{\theta}$, we first use the unit circle to compare the following areas as follows
$\frac{\sin\theta}{2} < \frac{\theta}{2} < \frac{\tan \theta}{2}$
or more simply,
$\sin \theta < \theta < \tan \theta$.
Taking reciprocals (which we can do since we are only consiering small $\theta > 0$) reverses the direction of the inequalities, so we have
$\frac{1}{\tan \theta} < \frac{1}{\theta} < \frac{1}{\sin \theta}$.
Then, multiplying by $\sin \theta$ (which is positive), and noting that $\frac{\sin \theta}{\tan \theta} = \cos \theta$, gives us
$\cos \theta < \frac{\sin \theta}{\theta} < 1$.
Our calculation only establishes this inequality chain for $0 < \theta < \frac{\pi}{2}$, but since all the terms are the same when we replace $\theta$ with $-\theta$, it is true for all nonzero $\theta \in \big(-\frac{\pi}{2}, \frac{\pi}{2}\big)$. Furthermore, our above argument shows that
$-\theta < \sin \theta < \theta$
for values of $\theta$ close to $0$, and hence
$\cos \theta = \sqrt{1 – \sin^{2} \theta} > \sqrt{1 – \theta^{2}} > 1 – \theta^{2}$,
so we have
$1 – \theta^{2} < \cos \theta < \frac{\sin \theta}{\theta} < 1$
for values of $\theta$ sufficiently close to $0$. Thus, by the Squeeze Theorem, since $\displaystyle{\lim_{\theta \to 0}} (1-\theta^{2}) = \displaystyle{\lim_{\theta \to 0}} 1 =1$, we conclude that $\displaystyle{\lim_{\theta \to 0}}\frac{\sin \theta}{\theta} = 1$.
Basically, I would like to know how we get $\sqrt{1 – \theta^{2}} > 1 – \theta^{2}$? Also, why can't we just use the Squeeze Theorem with $\cos \theta < \frac{\sin \theta}{\theta} < 1$, where $\theta \to 0$ to show that $\displaystyle{\lim_{\theta \to 0}} \frac{\sin \theta}{\theta}=1$? Lastly, I'm not completely sure why $-\theta < \sin \theta$.
Best Answer
Notice that if $0<x<1$ then $0<x^2<x<1$. For small $\theta, \sqrt{1-\theta^2} < 1$, so $\sqrt{1-\theta^2} > 1-\theta^2$.
I would assume that the author has not shown that $\lim_{\theta\to0} \cos\theta = 1$ yet, but has taught limits of polynomials like $1-\theta^2$. The side effect of how they are doing it here is that you get the cosine limit "for free".
Lastly, we already know that $0 < \sin\theta<\theta$, and it is obvious that this implies $-\theta <0<\sin\theta <\theta$, which is what you wanted.
This highlights a bigger problem, however. Earlier, you used the "fact" that $\theta$ was positive to justify taking reciprocals of an inequality. This usage means that you didn't prove $\lim_{x\to0} \frac{\sin x}{x} = 1$, but rather $\lim_{x\to0^+} \frac{\sin x}{x}$. To prove the original limit, you need to handle the case with negative $\theta$ as well.