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Math Help - Another Squeeze Theorem Question

  1. #1
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    Another Squeeze Theorem Question

    OK, I think that I have this Squeeze Theorem figured out. Can somebody please look at this to tell me if I do?

    Proove:  \displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}

    Let:
    f(x)=- \sqrt{x^3+x^2}
    g(x)= \sqrt{x^3+x^2}sin \frac{\pi}{2}
    h(x)=  \sqrt{x^3+x^2}

    I think that I can state: -1 \leq sin \frac{\pi}{x} \leq 1

    Then, multiplying each term by  \sqrt{x^3+x^2}:

    - \sqrt{x^3+x^2} \leq \sqrt{x^3+x^2}sin \frac{\pi}{2} \leq \sqrt{x^3+x^2}

    Since  \displaystyle \lim_{x\to 0 } f(x) = \displaystyle \lim_{x\to 0 } h(x) =0}, using Squeeze Theorem, I have proved that:

     \displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}

    Did I do this correctly? I appreciate anybody who takes a look at this for me.
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  2. #2
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    Quote Originally Posted by joatmon View Post
    OK, I think that I have this Squeeze Theorem figured out. Can somebody please look at this to tell me if I do?

    Proove:  \displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}

    Let:
    f(x)=- \sqrt{x^3+x^2}
    g(x)= \sqrt{x^3+x^2}sin \frac{\pi}{2}
    h(x)=  \sqrt{x^3+x^2}

    I think that I can state: -1 \leq sin \frac{\pi}{x} \leq 1

    Then, multiplying each term by  \sqrt{x^3+x^2}:

    - \sqrt{x^3+x^2} \leq \sqrt{x^3+x^2}sin \frac{\pi}{2} \leq \sqrt{x^3+x^2}

    Since  \displaystyle \lim_{x\to 0 } f(x) = \displaystyle \lim_{x\to 0 } h(x) =0}, using Squeeze Theorem, I have proved that:

     \displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}

    Did I do this correctly? I appreciate anybody who takes a look at this for me.
    Looks good.
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  3. #3
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    This had an unique graph.

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