Why does $\frac{|\sin\theta|}{2}<\frac{|\theta|}{2}<\frac{|\tan\theta|}{2}$ not imply that $1>\lim_{\theta\to 0}\frac{\sin\theta}{\theta}>1$?

Question

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?

Answer 1

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.

Answer 2

The proof is based on Squeeze Theorem which can be stated as

Squeeze Theorem: Let $f, g, h$ be real valued functions defined in a certain deleted neighborhood $I$ of $a$ and further let $$g(x) \leq f(x) \leq h(x) $$ for all $x\in I$. If $$\lim_{x\to a} g(x) =\lim_{x\to a} h(x) =L$$ then $\lim\limits_{x\to a}f(x) $ also exists and equals $L$.

Now here is another obvious fact which holds the key: if $a, b$ are real numbers such that $a<b$ then $a\leq b$. Why is this obvious? Because by definition $a\leq b$ if and only if "$a<b$ or $a=b$".

Thus if we have $$g(x) <f(x) <h(x) $$ then it automatically implies $$g(x) \leq f(x) \leq h(x) $$ and you can apply Squeeze theorem.

Understanding how order relations work is a must to have a grasp of calculus. The statement of Squeeze Theorem has hypotheses which take care of both the strong and weak inequalities and the conclusion holds in both cases.

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I later checked the video at Khan Academy linked in question which talks about the inequalities involved $$1\geq \frac{\sin\theta} {\theta} \geq \cos\theta\tag{1}$$ and then mentions "Squeeze Theorem" and then draws the conclusion $$\lim_{\theta\to 0}1\geq \lim_{\theta\to 0}\frac{\sin\theta}{\theta}\geq \lim_{\theta\to 0}\cos\theta\tag{2}$$ Well, Squeeze Theorem is not about equation $(1)$ implying equation $(2)$ but rather it says if $(1)$ holds and we also have $$\lim_{\theta\to 0}1=\lim_{\theta\to 0}\cos\theta\tag{3}$$ (which is true) then the limit $$\lim_{\theta\to 0}\frac{\sin\theta}{\theta} $$ also exists (that's the big conclusion we seek: the existence of limit) and it equals the limits in $(3)$ (that's the smaller part of the conclusion: value of limit).

If we use strict inequalities like $$1> \frac{\sin\theta} {\theta} > \cos\theta\tag{4}$$ then this is already included in $(1)$ and we have $(3)$ also so that the conclusion holds.

Thinking in this manner one should observe that the real surprise here is $$1>\cos\theta$$ and yet their limits are equal. But then the surprise gets diluted to obviousness because no one seriously has a doubt in $\lim_{\theta\to 0}\cos\theta=1$.