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!

Prove that lim x -> 0 sin (1/x) doesn't exist

  1. Apr 1, 2017 #1
    1. The problem statement, all variables and given/known data
    prove that
    [tex]

    \lim_{x\rightarrow 0} \sin \left( \frac{1}{x} \right)

    [/tex] doesn't exist.

    2. Relevant equations
    [tex]
    \lim_{x\rightarrow0}\sin\left(\frac{1}{x}\right)=\lim_{u\rightarrow\infty}\sin u
    [/tex]

    3. The attempt at a solution
    My strategy to solve this problem is to make [itex] u \rightarrow \infty [/itex] through different paths and show that the limits are different.
    So, first one is [itex] a_{u}=\frac{\pi}{2}+2\pi u,\,u\in\mathbb{N} [/itex]. Second one is [itex]
    b_{u}=u\pi,\,u\in\mathbb{N}[/itex].
    Both these have the same behavior as [itex] u \rightarrow \infty [/itex]:
    [tex]

    \lim_{u\rightarrow\infty}a_{u}=\lim_{u\rightarrow\infty}b_{u}=\infty.

    [/tex]
    For with first one,
    [tex]
    \begin{aligned}\lim_{u\rightarrow\infty}\sin a_{u} & =\lim_{u\rightarrow\infty}\sin\left(\frac{\pi}{2}+2\pi u\right)\\ & =\lim_{u\rightarrow\infty}\sin\left(\frac{\pi}{2}\right)\cos2\pi u+\sin2\pi u\cos\left(\frac{\pi}{2}\right)\\ & =\lim_{u\rightarrow\infty}\sin\left(\frac{\pi}{2}\right)\cos2\pi u+\lim_{u\rightarrow\infty}\sin2\pi u\cos\left(\frac{\pi}{2}\right)\\ & =\lim_{u\rightarrow\infty}\cos2\pi u+\lim_{u\rightarrow\infty}\sin2\pi u\cos\left(\frac{\pi}{2}\right). \end{aligned}
    [/tex]

    The limit [itex]
    \lim_{u\rightarrow\infty}\sin2\pi u\cos\left(\frac{\pi}{2}\right)
    [/itex] tends to 0, as [itex]
    \cos\left(\frac{\pi}{2}\right)
    [/itex] tends to 0, and [itex]
    \sin2\pi u
    [/itex] is limited both upper and lower bound.

    Therefore, [tex]
    \lim_{u\rightarrow\infty}\sin a_{u}=1.
    [/tex]

    Now, [tex]
    \lim_{u\rightarrow\infty}\sin b_{u}=\lim_{u\rightarrow\infty}\sin u\pi=0,
    [/tex] as, [itex] \forall u, u\pi [/itex] is an multiple of [itex]\pi[/itex], which [itex] \sin [/itex] is null.
    Then, [tex]
    \lim_{u\rightarrow\infty}\sin b_{u}=0.
    [/tex]
    Because of these two different results for different paths, it's true that [tex]
    \nexists\lim_{x\rightarrow0}\sin\frac{1}{x}.
    [/tex]

    Am I correct?
    Thanks in advance!
     
    Last edited: Apr 1, 2017
  2. jcsd
  3. Apr 1, 2017 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It looks okay, but could have been a lot simpler:

    ##\sin(\frac{\pi}{2} + 2\pi u) = \sin{\frac{\pi}{2}} = 1##

    Also, you could just let ##a_u = \frac{1}{\frac{\pi}{2} + 2\pi u}## and keep the limits going to ##0##.
     
  4. Apr 1, 2017 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You are overthinking this. The individual functions are constant, their limit is easy to find.
    Directly using sequences for the original function is better, because it avoids issues with the sign change at 0.
     
  5. Apr 1, 2017 #4
    You mean, I should use [itex]
    a_{u}=\frac{1}{\frac{\pi}{2}+2\pi u},\,u\in\mathbb{N}
    [/itex] and [itex]
    \lim_{x\rightarrow0}\sin\frac{1}{x}
    [/itex] since the beggining, correct?
     
  6. Apr 1, 2017 #5
     
  7. Apr 1, 2017 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I thought you were using the idea of finding two sequences that converge to 0 but for which the two sequences of function values converge to different values. That looks fine as an approach. Changing the limit to ##\infty## requires a bit more justification.
     
  8. Apr 1, 2017 #7
    Yes, that's exactly what's my initial approach was: two sequences converging to same value, but producing different limits with [itex] \sin \left( \frac{1}{x} \right) [/itex].
     
  9. Apr 1, 2017 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    One minor problem is that:

    [tex]
    \lim_{x\rightarrow0+}\sin\left(\frac{1}{x}\right)=\lim_{u\rightarrow\infty}\sin u
    [/tex]

    This is not fatal to your proof, but what you have is not quite right. It's better to stick with sequences going to 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Prove that lim x -> 0 sin (1/x) doesn't exist
  1. Lim log(sin(x))/log(x) (Replies: 11)

Loading...