as x approaches positive infinity

the square root (in parenthesis) of (x ^2 + ax) minus x

Thanks in advance.

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- Oct 26th 2009, 05:29 PMroflcopturFind the limit
as x approaches positive infinity

the square root (in parenthesis) of (x ^2 + ax) minus x

Thanks in advance. - Oct 26th 2009, 07:14 PMmr fantastic
- Oct 26th 2009, 07:14 PMScott H
Have you learned L'Hospital's rule? We can occasionally rewrite a limit of the form $\displaystyle \infty - \infty$ as a limit of the form $\displaystyle \frac{\infty}{\infty}$. In our case,

$\displaystyle \begin{aligned}

\lim_{\small x\rightarrow\infty}(\sqrt{x^2+ax}-x)&=\lim_{\small x\rightarrow\infty}\left(x\sqrt{1+\frac{a}{x}}-x\right)\\

&=\lim_{\small x\rightarrow\infty}\left(x\left(\sqrt{1+\frac{a}{x }}-1\right)\right)\\

&=\lim_{\small x\rightarrow\infty}\frac{\sqrt{1+\frac{a}{x}}-1}{\frac{1}{x}}.

\end{aligned}

$

Edit: Or you can use Mr. Fantastic's method, which doesn't require the rule. - Oct 26th 2009, 09:58 PMroflcoptur(:
I have learned both methods. I understand both of them now. Thanks!