<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] Another way to do it.
Considering that $x\to\infty$ $$\sqrt{x^2+8x}-\sqrt{x^2+7x}=x\left(\sqrt{1+\frac 8x}-\sqrt{1+\frac 7x} \right)$$ Now, using Taylor for small $y$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ make $y=\frac 8x$ in the first radical and $y=\frac 7x$ in the second radical to get $$\sqrt{x^2+8x}-\sqrt{x^2+7x}=x\left(\left(1+\frac{4}{x}-\frac{8}{x^2}+O\left(\frac{1}{x^3}\right)  \right)-\left(1+\frac{7}{2 x}-\frac{49}{8 x^2}+O\left(\frac{1}{x^3}\right)\right)\right)$$ After simplification $$\sqrt{x^2+8x}-\sqrt{x^2+7x}=\frac{1}{2}-\frac{15}{8 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.
Edit
Making the problem more general as Marty Cohen did in his answer, doing the same, we should get 
$$\sqrt{x^2+ax+c}-\sqrt{x^2+bx+d}=\frac{a-b}{2}+\frac{-a^2+4 b+c^2-4 d}{8 x}+O\left(\frac{1}{x^2}\right)$$
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