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Math Help - How do I solve this limit?

  1. #1
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    How do I solve this limit?

    \lim_{x\to\infty}\frac{sin(x^2+5)}{x^2}

    I know that sin(x^2+5) will oscillate between 1 and -1 and that \lim_{x\to\infty}\frac{1}{x^2} = 0

    Is this a case for using the squeeze theorem? How do I write out the answer correctly? Or is it just the limit of of the \frac{1}{x^2} function = 0 and the limit of the product of two functions is the product of the limits and 0*anything = 0 ?

    Thanks.
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  2. #2
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    Re: How do I solve this limit?

    Quote Originally Posted by terrorsquid View Post
    \lim_{x\to\infty}\frac{sin(x^2+5)}{x^2}

    I know that sin(x^2+5) will oscillate between 1 and -1 and that \lim_{x\to\infty}\frac{1}{x^2} = 0

    Is this a case for using the squeeze theorem? How do I write out the answer correctly? Or is it just the limit of of the \frac{1}{x^2} function = 0 and the limit of the product of two functions is the product of the limits and 0*anything = 0 ?

    Thanks.
    \displaystyle \sin{(x^2 + 5)} is bounded between -1 and 1, and \displaystyle x^2 increases without bound. Therefore, whatever the value of the numerator, dividing by \displaystyle x^2 when \displaystyle x is large makes the quotient very small.

    Therefore \displaystyle \lim_{x \to \infty}\frac{\sin{(x^2 + 5)}}{x^2} = 0
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    Re: How do I solve this limit?

    Maybe I over-complicated things because the question asked that I carefully justify every step of my argument. Thanks
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  4. #4
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    Re: How do I solve this limit?

    You could you the squeeze theorem: -\frac{1}{x^2}\le\frac{\sin(x^2+5)}{x^2}\le\frac{1}  {x^2}. It's also easy to show this by the definition of limit since \left|\frac{\sin(x^2+5)}{x^2}\right|\le\frac{1}{x^  2} and you can find from which point \frac{1}{x^2}<\varepsilon.
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    Re: How do I solve this limit?

    Quote Originally Posted by terrorsquid View Post
    Maybe I over-complicated things because the question asked that I carefully justify every step of my argument. Thanks
    Every step IS justified - it's more sophisticated to explain something in English than it is to explain something in symbols...
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  6. #6
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    Re: How do I solve this limit?

    Quote Originally Posted by terrorsquid View Post
    \lim_{x\to\infty}\frac{sin(x^2+5)}{x^2}

    I know that sin(x^2+5) will oscillate between 1 and -1 and that \lim_{x\to\infty}\frac{1}{x^2} = 0

    Is this a case for using the squeeze theorem? How do I write out the answer correctly? Or is it just the limit of of the \frac{1}{x^2} function = 0 and the limit of the product of two functions is the product of the limits and 0*anything = 0 ?

    Thanks.
    the trouble with the 0*anything = 0 "proof" is that the "anything" in this case, is undefined. so the "squeeze theroem" method makes more sense. or, you could appeal directly to the definition of a limit:

    let \epsilon > 0. choose N > \frac{1}{\sqrt{\epsilon}}.

    then, for x > N, \left|\frac{\sin(x^2+5)}{x^2}- 0\right| \leq \frac{1}{x^2} < \frac{1}{N^2} < \frac{1}{\frac{1}{(\sqrt{\epsilon})^2}} = \epsilon
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