# How does the squeeze theorem imply that $\lim_{\theta\to 0}\sin\theta=0$ and $\lim_{\theta\to 0}\cos\theta=1$?

I'm trying to understand how the sandwich/ squeeze theorem establishes that $$\lim_{\theta\to 0}\sin\theta = 0$$ and $$\lim_{\theta\to 0}\cos\theta = 1$$.

Mainly, i'm trying to understand this logic, rather than trying to answer an assignment question.

• Are you already equipped with the series representation of $\sin \theta$ and $\cos \theta$ and their approximation by Taylor polynomials? Because first you need an appropriate approximation of these functions near $0$ and then we could check how the sandwich trick works. Commented Feb 20, 2019 at 4:17

The idea behind sandwich/squeeze theorem is intuitive enough. If values of a function $$f$$ are lying (sandwiched) between those of $$g, h$$ and values of $$g, h$$ tend to a common limit then that of $$f$$ also tend to the same limit.

We have the fundamental inequality $$|\sin x|\leq |x|$$ for $$0<|x|<\pi/2$$ which is the same as $$-|x|\leq \sin x\leq |x|$$ for $$0<|x|<\pi/2$$ and since both $$-|x|$$ and $$|x|$$ tend to $$0$$ as $$x\to 0$$ it follows that $$\lim_{x\to 0}\sin x=0$$.

For $$\cos x$$ the situation is tricky and we need to use a bit of trigonometry. We have $$\cos x=1-2\sin^2(x/2)$$ and noting the earlier inequality for $$\sin$$ we get $$1-\frac{x^2}{2}\leq \cos x\leq 1$$ And by squeeze we get $$\lim_{x\to 0}\cos x=1$$.

Edit: I corrected some inaccuracies pointed out in the comments.

I try to explain the logic by solving the exercize. First of all, this result is also known as the two policemen theorem because if two policemen are escorting a person between them, and both officers approch arbitrary close to the cell, then the prisoner must also approch arbitrary close to it (notice that the policemen don't need to go to the cell; this means the functions need not to be defined in the limit point). This summarizes the main idea of the theorem.

Now, we know that when $$x$$ is sufficiently near to $$0$$ (i.e. the functions must be defined in a neighbourhood of such a point), $$-\lvert x\rvert\le \sin x\le \lvert x\rvert$$ and this holds for any $$x$$ sufficiently small (actually for any $$\lvert x\rvert$$ in $$]0,\pi/2[$$). Since limits preserve the linear order, we have that if limit of $$\sin x$$ exists, then $$\lim_{x\rightarrow0} -x\le \lim_{x\rightarrow0} \sin x \le \lim_{x\rightarrow0} x.$$ But the first and the last limits are $$0$$, hence limit of $$\sin x$$ exists and is $$0$$ because this function is eventually bounded both from above and below by functions converging to the same limit, so $$\sin x$$ can't go away.

Generally speaking, you need two functions/policemen who bound the function of which you want to compute the limit, from above and below respectively, in a neighbourhood of the limit point and they must have the same limit at that point.

• Strictly speaking, the officers don't actually have to enter the cell, they just have to approach arbitrarily close to it, and then the prisoner must also approach arbitrarily close to it. :-) Commented Feb 20, 2019 at 5:08
• There are some inaccuracies in this answer. The inequalities $-x \leq \sin \, x\leq x$ can hold only for $x \geq 0$ so one can only take right hand limits. Commented Feb 20, 2019 at 5:16
• @BrianTung I was so drunk last night that I don't remember exactly what happened with the officers ;-) Commented Feb 20, 2019 at 9:19
• @KaviRamaMurthy I apologize for the errors and thanks for the mistakes you pointed out. Commented Feb 20, 2019 at 9:19