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Limit using Sandwich Theorem

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\lim_{x \to 0} \sqrt{x^3 + x^2}\sin\frac{\pi}{x}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    We know that [itex] -1 \leq \sin\frac{\pi}{x} \leq 1 [/itex]
    [itex]\Leftrightarrow -\sqrt{x^3 + x^2} \leq \sqrt{x^3 + x^2}\sin\frac{\pi}{x} \leq \sqrt{x^3 + x^2}[/itex] since [itex]\sqrt{y} \geq 0\forall y\in\mathbb{P}\cup\{0\}[/itex]
    Now, [itex]\lim_{x \to 0} -\sqrt{x^3 + x^2} = 0 = \lim_{x \to 0} \sqrt{x^3 + x^2}[/itex]
    [itex] \therefore \lim_{x \to 0} \sqrt{x^3 + x^2}\sin\frac{\pi}{x} = 0 [/itex] by Sandwich Theorem.
     
    Last edited: Aug 21, 2011
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  3. Aug 21, 2011 #2

    vela

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    What's your question?
     
  4. Aug 21, 2011 #3
    I want to know if it is correct.

    What worries me is if I can state that [itex] -1 \leq \sin{\frac{\pi}{x}} \leq 1 [/itex] since x is approaching 0, and right there [itex] \frac{\pi}{x} [/itex] is undefined. Thanks in advance.
     
  5. Aug 21, 2011 #4

    SammyS

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    What you can say is -1 ≤ sin(θ) ≤ 1 for all real θ.

    Therefore, [itex]-1 \leq \sin{\frac{\pi}{x}} \leq 1[/itex] provided that x ≠ 0 .

    Since x approaches zero for your limit, x is not equal to zero, so you're fine.
     
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