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Limit of xsin(1/x) as x→0

  1. Dec 17, 2011 #1
    1. The problem statement, all variables and given/known data
    [itex]\displaystyle\lim_{x\rightarrow 0}~~ {xsin(1/x)}[/itex]

    3. The attempt at a solution
    I've attempted to solve this limit two different ways and get different answers.

    Attempt #1:
    If [itex](x_n)[/itex] is a sequence and xn→0 then because sin(1/xn) is bounded xsin(1/x)→0. So [itex]\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}[/itex]=0

    Attempt #2:
    [itex]\displaystyle\lim_{x\rightarrow 0} ~~{xsin(1/x)}[/itex] = [itex]\displaystyle\lim_{x\rightarrow 0}~~ {\frac{(1/x)(x)(sin(1/x))}{(1/x)}}[/itex] = [itex](1/x)(x) ~~\displaystyle\lim_{x\rightarrow 0} ~~ {\frac{sin(1/x)}{(1/x)}}[/itex]= (1)(1) = 1

    Can you guys tell me which is correct, and what is my error in the incorrect attempt? Thanks!
  2. jcsd
  3. Dec 17, 2011 #2
    Would you happen to be familiar with what a taylor series is? Because if you know the series function for sine then the answer would be very clear,

    nevertheless here is another way to look at it:

    What happens to expression 1/Xn as Xn approaches zero...
  4. Dec 17, 2011 #3
    Follow up to that what happens then to sine (1/Xn) as Xn approaches zero
  5. Dec 17, 2011 #4
    as Xn→0, 1/Xn approaches ±∞. The limit as it approaches from the left does not equal the limit as it approaches from the right, so the limit of 1/Xn as Xn approaches 0 does not exist. So the limit of sin(1/xn) also doesn't exist?

    Can you pinpoint exactly where I went wrong in my attempts or is this question simply a special case?
  6. Dec 17, 2011 #5
    don't think it does not exist JUST YET...

    What is the range of sin(x)?

    What happens to the actual graph as sin(1/x) approaches zero from -1 to 0 and from 1 to zero?
  7. Dec 17, 2011 #6
    The limit as x approaches zero of x * sin(1/x)

    is taking the limit as x approaches zero and multiplying it with the limit as sin(1/x) approaches zero...

    So we're trying to find out what happens to the behavior as it gets closer to zero...

    Keep in mind that sin of anything is restricted to a range of [-1, 1]

    One additional clue, The function sin(1/x) oscillates increasingly faster as x approaches zero so at zero itself its undefined because it could be any number in [-1, 1]...

    But what is the value of x as it approaches zero? and what happens if you multiply the value of sin(1/x) with x? Does it converge back to an answer from being undefined or is it still undefined?


    Recalculate the Limit as x approaches 0 for sin(1/x)/(1/x) and tell me what answer you get
  8. Dec 17, 2011 #7
    The range of sin x is [-1,1], so the range of sin (1/x) is also [-1,1]. Because the limit of x as x→0 = 0, multiplying this by sin(1/x) will give us 0 (because range of sin(1/x) is bounded). So I would think that the limit of (x)(sin1/x) as x→0 would equal 0.

    So the limit of (sin1/x)/(1/x) is not defined, because sin (1/x) stays between [-1,1] but (1/x) will increase/decrease to ±∞ as x approaches 0. Because the bottom is increasing/decreasing to ±∞, it the limit of (sin 1/x)/(1/x) is also 0. Is this now correct?

    I'm curious as to why I can't use the property that the limit of (sin x)/x as x→0 = 1 for this example, plugging in (1/x) for x?
  9. Dec 17, 2011 #8
    Why can't you?
  10. Dec 17, 2011 #9


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    Use the sandwich or squeeze theorem.

    First you say the limit isn't defined, and then you say it's also 0. Which is it? :wink:

    It's because 1/x goes to ∞ as x goes to 0, so you have
    [tex]\lim_{x \to 0} \frac{\sin (1/x)}{1/x} = \lim_{u \to \infty} \frac{\sin u}{u}[/tex]The second limit would equal 1 if u→0, but that's not what you have.
  11. Dec 17, 2011 #10

    Could I also use the domination principle? Because sin(1/x) is between [-1,1] then |x*sin(1/x)| < |x| where x→0 so |x|→0 and so |x||sin(1/x)| also converges to 0?

    Thank you for the explanation!!
  12. Dec 17, 2011 #11


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    Another mistake in your second attempt is that you can't simply pull the factors containing x out of the limit since you're taking the limit as x→0. You can only pull constant factors out.
  13. Dec 17, 2011 #12


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    As vela has already suggested, use the squeeze theorem.

    -|x| ≤ x sin(1/x) ≤ |x|
  14. Dec 17, 2011 #13


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    This is essentially an application of the squeeze theorem. You have [itex]0 \le |x||\sin(1/x)| \le |x|[/itex] (note it should be ≤, not <, because sin(1/x) can equal 1), and because both the left and right go to 0, the middle must also go to 0.

    So you've shown that
    [tex]\lim_{x \to 0}\ |x\sin(1/x)| = 0[/tex]You still need to show why this implies
    [tex]\lim_{x \to 0}\ x\sin(1/x) = 0[/tex]
  15. Dec 17, 2011 #14
    Is this simply because |x| = 0 then x=0? So [itex]\lim_{x \to 0}\ x\sin(1/x)[/itex] must equal 0 because its absolute value limit is 0?
  16. Dec 17, 2011 #15
    [tex]\lim_{x\to 0}{\left|x \sin{\frac{1}{x}}\right|}=0[/tex]
    [tex]\forall \varepsilon>0\hskip10pt \exists \delta>0 \hskip10pt\ 0<|x-0|<\delta \hskip10pt\ \left|\left|x\sin{\frac{1}{x}}\right|-0\right|<\varepsilon[/tex]
    But, [tex]\left|x\sin{\frac{1}{x}}\right|=\left|\left|x\sin{\frac{1}{x}}\right|\right|<\varepsilon[/tex]
    so we can say
    [tex]\lim_{x\to 0}{x\sin{\frac{1}{x}}}=0[/tex]
    Is this OK?
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