Limit of ($\sqrt{x^2+8x}-\sqrt{x^2+7x}$) as $x$ approaches infinity

Question

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?

You could also, after multiplying and dividing by the conjugate, use L'Hôpital's rule to analyze the denominator. You find that this limit behaves as the inverse of its conjugate, which tends to one-half as $x$ goes to infinity. A more general analysis may also be conceived of this way, if you make some assumptions about the coefficients. — KR136, May 29 '16 at 04:50
Try http://math.stackexchange.com/questions/564132/having-trouble-with-a-lim-infinity-infinity/564234#564234 — lab bhattacharjee, May 29 '16 at 06:28
Duplicate: $\displaystyle\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$ (with $a=8$ and $b=7$). (Found using Approach0.xyz) — Workaholic, Nov 02 '16 at 08:13

score 4 · Answer 1 · answered May 29 '16 at 03:25

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 $x<0$

score 1 · Answer 2 · edited May 29 '16 at 04:30

1

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} $.

edited May 29 '16 at 04:30

answered May 29 '16 at 03:57

marty cohen

107,799

2

You forgot to divide by 2 ( $=\sqrt{1}+\sqrt{1}$ ) when calculating the limit. – MasB May 29 '16 at 04:02
Thanks. Another careless error. Oh well. – marty cohen May 29 '16 at 04:38

score 1 · Answer 3 · edited May 29 '16 at 04:34

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

$$\sqrt{x^2+ax+c}-\sqrt{x^2+bx+d}=\frac{a-b}{2}+\frac{-a^2+4 b+c^2-4 d}{8 x}+O\left(\frac{1}{x^2}\right)$$

You mean ${a-b\over 2}$? I edited. I think that for any limit one should try to go through the asymptotic series to see the way the limit is approached — marwalix, May 29 '16 at 04:33
@marwalix. I thought I typed that ... but I certainly missed something ! Thanks for editing. Cheers. — Claude Leibovici, May 29 '16 at 04:39

Limit of ($\sqrt{x^2+8x}-\sqrt{x^2+7x}$) as $x$ approaches infinity

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