Silly Squeeze Theorem Question lim (cos(x))

[Saladsamurai]

Homework Statement

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<br /> <br /> \lim_{\theta\rightarrow 0}\cos\theta = 1 ?

Thanks!

 

[JG89]

<br /> \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

 

[Saladsamurai]

<br /> \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

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

 

[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| &lt; \epsilon whenever |\theta| &lt; \delta.

Note that |1 - cos\theta| = |(1 - cos\theta) - 0| &lt; \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.

 

[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).

 

[slider142]

<br /> \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

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.

 

[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.

 

[slider142]

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.

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.

 

[JG89]

I stand corrected. Thanks for the counter example!

 

[JG89]

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

 

[slider142]

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

Nice argument!