# Using the squeeze theorem to show that $\text{lim }_{\theta \rightarrow 0} \text{sin }\left(\theta\right) =0$

I'm a TA at my university and recently encountered this problem, I know that the answer must be quite simple, but I seem to be at a blank at the moment. Could someone please help me with this? Any help is appreciated.

Using the squeeze theorem to show that $\text{lim }_{\theta \rightarrow 0} \text{sin }\left(\theta\right) =0$ and from this conclude that $\text{lim }_{\theta \rightarrow 0} \text{cos }\left(\theta\right) =1$

What about simply using continuity of $\;\sin x\;$ ? :

$$\lim_{x\to 0}\sin x=\sin\left(\lim_{x\to 0}x\right)=\sin 0=0$$

If you know that $\;-x\le\sin x\le x\;$ (and we only care when $\;x>0\;$ is close to zero), then the squezze theorem gives us the result at once.

And using trigonometry:

$$\cos^2 x=1-\sin^2x\implies \lim_{x\to 0}\cos^2 x=\lim_{x\to 0}(1-\sin^2x)=1-0=1$$

Let us approach $0$ from the right. Then we know that $\sin \theta >0$ also geometrical arguments using the circle as are done to prove that $\lim \frac{\sin \theta}{\theta }=1$ yield that $\sin \theta< \theta$. Now you can squeeze it.

• Near $0$ is it safe to assume $1-x \leq \text{cos }(x) \leq 1+x$? – Pablo Mar 3 '15 at 10:00
• @Pablo yes you can assume that since the power series for $\cos = 1-\frac{x^2}{2}+...$. $\cos(x) <1$ always and $x$ close to $0$ implies $x^2/2<x/2<x$. So $1-\frac{x^2}{2}>1-x$ This is not a complete justification you will have to work the details – happymath Mar 3 '15 at 10:05
• Your argument suggests that "since $\lim\frac{\theta}{\sin \theta} = 1$ we have $\theta<\sin\theta$". – AD. Mar 3 '15 at 10:19
• @A.D. I am sorry if I am misleading but I am not using the fact that the limit is $1$ rather I am using the method in the proof to get sin $\theta<\theta$ – happymath Mar 3 '15 at 10:22