1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Silly Squeeze Theorem Question lim (cos(x))

  1. Aug 14, 2009 #1
    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[itex]\theta[/itex], [itex]0\le 1-\cos\theta\le |\theta|[/itex] for all theta. and we have

    [tex]\lim_{\theta\rightarrow 0}(1-\cos\theta)=0[/tex] or

    [tex]\lim_{\theta\rightarrow 0}\cos\theta = 1[/tex]

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

    [tex]\lim_{\theta\rightarrow 0}(1-\cos\theta)=0 \Rightarrow

    \lim_{\theta\rightarrow 0}\cos\theta = 1[/tex] ?

    Thanks! :smile:
     
  2. jcsd
  3. Aug 14, 2009 #2
    [tex]
    \lim_{\theta\rightarrow 0}(1-\cos\theta)= \lim_{\theta \rightarrow 0} 1 - \lim_{\theta \rightarrow 0} cos\theta [/tex] and since [tex] \lim_{\theta \rightarrow 0} (1 - cos\theta) = 0 [/tex], we have [tex] \lim_{\theta \rightarrow 0} cos\theta = \lim_{\theta \rightarrow 0} 1 = 1 [/tex]
     
  4. Aug 14, 2009 #3
    Awwww man!!! Oldest trick in the book! Can't belive I didn't see that! Thanks JG89! :smile:
     
  5. Aug 14, 2009 #4
    Here's another way to look at it:

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

    Note that [tex] |1 - cos\theta| = |(1 - cos\theta) - 0| < \epsilon [/tex]. So this statement says that for theta approaching 0, [tex] 1 - cos\theta [/tex] goes to 0, or since |1 - cos(theta) - 0| = |1 - cos(theta)|, that cos(theta) goes to 1.
     
  6. Aug 14, 2009 #5
    If [tex]\lim_{x\rightarrow a} f(x) = L[/tex] and [tex]\lim_{x\rightarrow a} g(x) = M[/tex], then [tex]\lim_{x\rightarrow a} (f + g)(x) = L + M[/tex]. It is then a simple corollary that [tex]\lim_{x\rightarrow a} f(x) = L[/tex] if and only if [tex]\lim_{x\rightarrow a}(f(x) - L) = 0[/tex]. (The same theorem is used to prove both ways; apply it creatively).
     
  7. Aug 14, 2009 #6
    Note that the passage from the first equation to the second is only valid in if you already know that the [tex]\lim_{x\rightarrow 0} \cos x[/tex] exists, something you would have to argue or show separately.
     
  8. Aug 14, 2009 #7
    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.
     
  9. Aug 14, 2009 #8
    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.
     
  10. Aug 14, 2009 #9
    I stand corrected. Thanks for the counter example!
     
  11. Aug 14, 2009 #10
    Though in my defense, if for every positive epsilon there exists a positive delta such that [tex] |1 - cosx| < \epsilon [/tex] whenever [tex] |x| < \delta [/tex], then since cos(0) = 1, then we have [tex] |1 - cosx| = | cos0 - cosx| < \epsilon [/tex] if [tex] |x| < \delta [/tex], which means cos(x) is continuous at x = 0, so it must possess a limit there.
     
  12. Aug 15, 2009 #11
    Nice argument!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Silly Squeeze Theorem Question lim (cos(x))
  1. Squeeze theorem (Replies: 2)

Loading...