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Math Help - If x goes very large, say infinite, what does this equation will look like

  1. #1
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    If x goes very large, say infinite, what does this equation will look like

    when x goes very large, say infinite, what does this equation will look like?

    H is a constant

    1) 1/(1+(x/H)^2)^(1/2))

    2) 1/(1+(H/x)^2)^(1/2))

    The answers are

    1) 1/x^2

    2) 1/x

    But I am not too sure how to get there. Thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by oxxiissiixxo View Post
    when x goes very large, say infinite, what does this equation will look like?

    H is a constant

    1) 1/(1+(x/H)^2)^(1/2))

    2) 1/(1+(H/x)^2)^(1/2))

    The answers are

    1) 1/x^2

    2) 1/x

    But I am not too sure how to get there. Thank you

    Did you mean?:

    \dfrac{1}{\sqrt{1+(x/H)^2}} for 1) ?
    Last edited by FernandoRevilla; February 28th 2011 at 07:55 AM.
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  3. #3
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    1) \displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{x}{H}\right)^2}}

    As x grows larger, the denominator of the fraction grows larger. In this regard, the limit is similar to: \displaystyle \lim_{t\to\infty}\frac{1}{t}, which is 0.

    2) \displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{H}{x}\right)^2}}

    As x grows larger, \frac{H}{x} approaches 0. Therefore,

    \displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{H}{x}\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(0\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+0}} = \lim_{x\to\infty} \frac{1}{\sqrt{1}} = \lim_{x\to\infty} \frac{1}{1} = \lim_{x\to\infty} 1 = 1
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