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?
 A: Hint: $\sqrt{x^2+8x}-\sqrt{x^2+7x}={({\sqrt{x^2+8x}-\sqrt{x^2+7x})(\sqrt{x^2+8x}+\sqrt{x^2+7x})}\over{\sqrt{x^2+8x}+\sqrt{x^2+7x}}}={x\over {\mid x\mid\sqrt{1+8/x}+\mid x\mid \sqrt{1+7/x}}}$
${1\over {\sqrt{1+8/x}+ \sqrt{1+7/x}}}$ if $x>0$ and
${-1\over {\sqrt{1+8/x}+ \sqrt{1+7/x}}}$ if $x<0$
A: 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}
$.
A: 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 
$$\sqrt{x^2+ax+c}-\sqrt{x^2+bx+d}=\frac{a-b}{2}+\frac{-a^2+4 b+c^2-4 d}{8 x}+O\left(\frac{1}{x^2}\right)$$
