I was watching this proof of the equality $$\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1$$
The author says about the following areas that
red area <= yellow area <= blue area. Which leads to the following inequality:
$$\frac{|\sin\theta|}{2} \le \frac{|\theta|}{2} \le \frac{|\tan\theta|}{2}$$
and in the end proofs the theorem.
$$1 \ge \lim_{\theta\to 0} \frac{\sin \theta}{\theta} \ge 1 $$
I noticed that the statement red area < yellow area < blue area about the areas is also true and in fact more accurate. But this would lead to the following:
$$\frac{|\sin\theta|}{2} \lt \frac{|\theta|}{2} \lt \frac{|\tan\theta|}{2}$$
…
$$1 \gt \lim_{\theta\to 0} \frac{\sin \theta}{\theta} \gt 1 $$
Obviously that cannot be true.
Have I just broken the proof?
Best Answer
Good first question! Even if $f(x)<M$ for some value $M$ and for every $x\neq x_0$ in the domain of $f$, you still can't conclude that $\lim\limits_{x\to x_0}f(x)<M$. The most you can say is that $\lim\limits_{x\to x_0}f(x)\leq M$. For example, if $f(x)=1-x^2$, then $f(x)<1$ for all $x\neq0$, but $\lim\limits_{x\to0}1-x^2=1$.
In your case, you have: $$\cos\theta<\frac{\sin\theta}{\theta}<1$$ for every $\theta\neq0$ (at least in a neighborhood of $\theta=0$), so in the limit you have: $$1\leq\lim_{\theta\to0}\frac{\sin\theta}{\theta}\leq 1$$ which is true! So the proof is not broken after all.