Find the limit

• Oct 26th 2009, 05:29 PM
roflcoptur
Find the limit
as x approaches positive infinity
the square root (in parenthesis) of (x ^2 + ax) minus x

• Oct 26th 2009, 07:14 PM
mr fantastic
Quote:

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

Multiply by $\displaystyle \frac{\sqrt{x^2 + ax} + x}{\sqrt{x^2 + ax} + x}$, 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.
• Oct 26th 2009, 07:14 PM
Scott 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 PM
roflcoptur
(:
I have learned both methods. I understand both of them now. Thanks!