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Continuity of sine function

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Using the inequality [tex]|\sin(x)| < |x|[/tex] for [tex] 0 < |x| < \frac{\pi}{2}[/tex], prove that the sine function is continuous at 0.

    2. Relevant equations

    Definition of continuity: A function f: R -> R is continuous at a point [tex]x0 \in R[/tex], if for any [tex]\epsilon > 0[/tex], there esists a [tex]\delta(\epsilon; x0)[/tex] such that if [tex]|x - x0| \leq \delta(\epsilon; x0)[/tex], then [tex]|f(x) - f(x0)| \leq \epsilon[/tex].

    3. The attempt at a solution
    To be honest, I am not really sure about this. I know that in order to prove continuity at a point, I can take the limit of the function at that point and see if it's equal to the value of the function. With this method, though, I am not using the inequality given, and therefore I cannot think of a way to do this. In class we were given no examples, so I am really lost here. I tried using the definition of continuity, but I did not manage to get anywhere that made sense.

    I am sure this cannot be hard and I am probably missing something obvious. If anyone could help I'd truly appreciate it. Thank you.
  2. jcsd
  3. Jan 18, 2010 #2


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    Alright, are you familiar with the squeeze theorem, because you're essentially asked to provide a formal illustration of this theorem. My suggestion is let [itex]\delta = \varepsilon[/itex] and then show that if [itex]0 < |x| < \delta[/itex] then [itex]|\sin{(x)}| < \varepsilon[/itex].
  4. Jan 18, 2010 #3
    I am not familiar with the squeeze theorem, but I will look into it. Thanks for the tip, I'll let you know if I get anywhere.
  5. Jan 18, 2010 #4


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    Hint: [itex]|\sin x| \le |x|[/itex].
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