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Math Help - Find the limit

  1. #1
    Newbie roflcoptur's Avatar
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    Question Find the limit

    as x approaches positive infinity
    the square root (in parenthesis) of (x ^2 + ax) minus x

    Thanks in advance.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by roflcoptur View Post
    as x approaches positive infinity
    the square root (in parenthesis) of (x ^2 + ax) minus x

    Thanks in advance.
    Multiply by \frac{\sqrt{x^2 + ax} + x}{\sqrt{x^2 + ax} + x}, simplify the numerator, divide numerator by x and then take the limit.

    If you need more help, please post your work and say where you get stuck.
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  3. #3
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    Have you learned L'Hospital's rule? We can occasionally rewrite a limit of the form \infty - \infty as a limit of the form \frac{\infty}{\infty}. In our case,

    \begin{aligned}<br />
\lim_{\small x\rightarrow\infty}(\sqrt{x^2+ax}-x)&=\lim_{\small x\rightarrow\infty}\left(x\sqrt{1+\frac{a}{x}}-x\right)\\<br />
&=\lim_{\small x\rightarrow\infty}\left(x\left(\sqrt{1+\frac{a}{x  }}-1\right)\right)\\<br />
&=\lim_{\small x\rightarrow\infty}\frac{\sqrt{1+\frac{a}{x}}-1}{\frac{1}{x}}.<br />
\end{aligned}<br />

    Edit: Or you can use Mr. Fantastic's method, which doesn't require the rule.
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  4. #4
    Newbie roflcoptur's Avatar
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    (:

    I have learned both methods. I understand both of them now. Thanks!
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