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Lim x->0 (sin x / x) =1 contradiction?

  1. Nov 17, 2011 #1
    lim x->0 (sin x / x) =1......contradiction?

    sin(x)/x =1 (limit x to 0)
    this is an identity proved by using geometry and squeeze theorem ...right?

    now today i came across another question and doing it my way ....gives me two answers;)
    the question is limit x-->0 of [ x*sin(1/x)]

    my first approach was using the above identity by rewriting the question as follows:
    sin(1/x)/(1/x)....it means the same thing and is now in the form so that we can use the identity....so limiting x->0 must give us 1 according to the identity

    now i thought to use the squeeze theorem as shown below:
    now x goes to 0 so:
    so x*sin(1/x)=0 as x goes to 0

    now which method which is correct ?
  2. jcsd
  3. Nov 17, 2011 #2


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    Re: lim x->0 (sin x / x) =1......contradiction?

    sin(1/x)/(1/x) as x goes to 0 would be the same as sin(x)/x as x goes to infinity, not 0.
  4. Nov 17, 2011 #3
    Re: lim x->0 (sin x / x) =1......contradiction?

    so it will be of the form sin(x)/x
    applying L'hopitals rule :
    = cos(x)
    as you said x goes to infinity so what will be cos(∞)?
  5. Nov 17, 2011 #4

    D H

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    Re: lim x->0 (sin x / x) =1......contradiction?

    It's undefined, and all that that means is that you can't use l'Hôpital's rule here.

    You can use the squeeze theorem on sin(x)/x as x→∞. sin(x) is bounded from above by +1, from below by -1. Given that, what does the squeeze theorem say about sin(x)/x as x→∞?
  6. Nov 17, 2011 #5
    Re: lim x->0 (sin x / x) =1......contradiction?

    ok ....got your point....its 0
    thx a lot
  7. Nov 17, 2011 #6


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    Re: lim x->0 (sin x / x) =1......contradiction?

    L'Hopital's rule doesn't apply, since |sin(x)| is bounded while the denominator x becomes infinite.
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