<s>[INST] Limit of ($\sqrt{x^2+8x}-\sqrt{x^2+7x}$) as $x$ approaches infinity I've been stuck on this one problem for 3 days now, I don't know how to proceed. Any help would be appreciated.
The problem is asking for the $$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x}) $$ 
Every time I attempt this problem, I can never get rid of infinity in the numerator without making the denominator zero. How can I do this problem?
 [/INST] You want
$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x})
$.
For reals $a$ and $b$,
$\begin{array}\\
\sqrt{x^2+ax}-\sqrt{x^2+bx}
&=(\sqrt{x^2+ax}-\sqrt{x^2+bx})\dfrac{\sqrt{x^2+ax}+\sqrt{x^2+bx}}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{(\sqrt{x^2+ax}-\sqrt{x^2+bx})(\sqrt{x^2+ax}+\sqrt{x^2+bx})}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{ax-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{x(a-b)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{a-b}{\sqrt{1+a/x}+\sqrt{1+b/x}}\\
&\to \dfrac{a-b}{2} \text{ as } x \to \infty\\
\end{array}
$
The result is the same
if the expression is
$\sqrt{x^2+ax+c}-\sqrt{x^2+bx+d}
$.
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