# Silly Squeeze Theorem Question lim (cos(x))

1. Aug 14, 2009

1. The problem statement, all variables and given/known data

Hoowwwwdddyyyyyy!

I am reviewing some Calculus and was going over the Sandwich Theorem. It's delicious.

An example that they give is: From the definition of cos$\theta$, $0\le 1-\cos\theta\le |\theta|$ for all theta. and we have

$$\lim_{\theta\rightarrow 0}(1-\cos\theta)=0$$ or

$$\lim_{\theta\rightarrow 0}\cos\theta = 1$$

I am having trouble seeing the transition in the last two steps. How does

$$\lim_{\theta\rightarrow 0}(1-\cos\theta)=0 \Rightarrow \lim_{\theta\rightarrow 0}\cos\theta = 1$$ ?

Thanks!

2. Aug 14, 2009

### JG89

$$\lim_{\theta\rightarrow 0}(1-\cos\theta)= \lim_{\theta \rightarrow 0} 1 - \lim_{\theta \rightarrow 0} cos\theta$$ and since $$\lim_{\theta \rightarrow 0} (1 - cos\theta) = 0$$, we have $$\lim_{\theta \rightarrow 0} cos\theta = \lim_{\theta \rightarrow 0} 1 = 1$$

3. Aug 14, 2009

Awwww man!!! Oldest trick in the book! Can't belive I didn't see that! Thanks JG89!

4. Aug 14, 2009

### JG89

Here's another way to look at it:

If $$\lim_{\theta \rightarrow 0} (1 - cos\theta) = 0$$ then for all positive epsilon there exists a positive delta such that $$|1 - cos\theta| < \epsilon$$ whenever $$|\theta| < \delta$$.

Note that $$|1 - cos\theta| = |(1 - cos\theta) - 0| < \epsilon$$. So this statement says that for theta approaching 0, $$1 - cos\theta$$ goes to 0, or since |1 - cos(theta) - 0| = |1 - cos(theta)|, that cos(theta) goes to 1.

5. Aug 14, 2009

### slider142

If $$\lim_{x\rightarrow a} f(x) = L$$ and $$\lim_{x\rightarrow a} g(x) = M$$, then $$\lim_{x\rightarrow a} (f + g)(x) = L + M$$. It is then a simple corollary that $$\lim_{x\rightarrow a} f(x) = L$$ if and only if $$\lim_{x\rightarrow a}(f(x) - L) = 0$$. (The same theorem is used to prove both ways; apply it creatively).

6. Aug 14, 2009

### slider142

Note that the passage from the first equation to the second is only valid in if you already know that the $$\lim_{x\rightarrow 0} \cos x$$ exists, something you would have to argue or show separately.

7. Aug 14, 2009

### JG89

Well if the limit as x approaches 0 of 1 - cos(x) exists, something which is being assumed in the OP's question, of course the limit of cos(x) exists.

8. Aug 14, 2009

### slider142

This is not true! Consider f(x) to be -1 for x in (-oo, 0) and 1 for x in [0, oo) and let g(x) be 1 for x in (-oo, 0] and -1 for x in (0, oo). Then as x approaches 0, (f + g)(x) approaches 0 as well, but individually the limits of f and g do not even exist at 0!
To bridge the step in your theorem, you must either know or prove beforehand that cos(x) is continuous at 0 or that it at least has a limit there.

9. Aug 14, 2009

### JG89

I stand corrected. Thanks for the counter example!

10. Aug 14, 2009

### JG89

Though in my defense, if for every positive epsilon there exists a positive delta such that $$|1 - cosx| < \epsilon$$ whenever $$|x| < \delta$$, then since cos(0) = 1, then we have $$|1 - cosx| = | cos0 - cosx| < \epsilon$$ if $$|x| < \delta$$, which means cos(x) is continuous at x = 0, so it must possess a limit there.

11. Aug 15, 2009

### slider142

Nice argument!