Limit of difference of two square roots

I need to find the limit, not sure what to do. $\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$

I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.

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Did you try multiplying by $\displaystyle\frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$? – t.b. Sep 6 '11 at 22:33
@Jordan: This is precisely the same type of question that you asked 51 minutes ago. Did you try to apply the same techniques? How long have you thought about this problem? What have you tried? What are the "weird numbers that don't seem to help"? – JavaMan Sep 6 '11 at 22:33
I multiplied by the conjugate and I got (1+a/x) - (1+b/x) and I dont think that helps. – user138246 Sep 6 '11 at 22:36
@Jordan If by that you mean the last expression in Edvin's answer, then that does help. $\lim_{x \to \infty} a/x$ is about as good as it gets. – Dylan Moreland Sep 6 '11 at 22:54

Applying the formula $x^2-y^2=(x-y)(x+y)$ we get

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

now you can take the limit.

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Is there any way to do this without knowing that trick? – user138246 Sep 6 '11 at 22:41
You could use the binomial expansion: write $\sqrt{x^2+ax}$ as $x\sqrt{1+{a\over x}}$, expand the square root out as $1+{a\over 2x}-{a^2\over 8x^2}+\ldots$, multiply back through by $x$ and subtract the corresponding calculation for the $\sqrt{x^2+bx}$ term. – Steven Stadnicki Sep 6 '11 at 22:46
This is all too complicated for me and I would never be able to reproduce this without a lot of practice. – user138246 Sep 6 '11 at 22:47
You have now in less than an hour asked two questions where the exact same technique yields the answer immediately. That is about the clearest possible indication that it is not "unique to each" problem. – Henning Makholm Sep 6 '11 at 23:14
You must practise more and definitely be more patient. – Edvin Goey Sep 6 '11 at 23:23

Alternative Method:
One may write the limit as, $$\left(\lim_{x\to\infty}\sqrt{x^2+ax}-x\right)-\left(\lim_{x\to\infty}\sqrt{x^2‌​+bx}-x\right)\!\, \!,$$ and then use the standard techniques for evaluating limits of the form, $$\lim_{x\to\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x,$$ to conclude that $\lim\limits_{x\to\infty}\sqrt{x^2+ax}-x=\tfrac a2$, and similarly $\lim\limits_{x\to\infty}\sqrt{x^2+bx}-x=\tfrac b2$, hence the limit is $\tfrac a2-\tfrac b2$.

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This answer may be a bit longer, but I've tried to include every step in the process. $$\label{eq1} \begin{split} \lim_{x \to \infty} \sqrt{x^2 + ax} - \sqrt{x^2 + bx} & = \lim_{x \to \infty} \left(\sqrt{x^2 + ax} - \sqrt{x^2 + bx}\right)\left(\frac{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}\right) \\ \\ & = \lim_{x \to \infty}\frac{(x^2 + ax) - (x^2 + bx)}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \\ \\ & = \lim_{x \to \infty}\frac{ax-bx}{\sqrt{x^2\left(1 + \frac{a}{x}\right)} + \sqrt{x^2\left(1 + \frac{b}{x}\right)}} \\ \\ & = \lim_{x \to \infty}\frac{ax-bx}{|x|\sqrt{1 + \frac{a}{x}} + |x|\sqrt{1 + \frac{b}{x}}} \\ \\ & = \lim_{x \to \infty}\frac{x(a-b)}{|x|\left(\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}\right)} \\ \\ &\text{Recall:} \ \ \ |x| = x \ \ \ \text{as} \ \ \ x \to \infty \\ \\ & = \lim_{x \to \infty}\frac{x(a-b)}{x\left(\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}\right)} \\ \\ & = \lim_{x \to \infty}\frac{a-b}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} \\ \\ & = \frac{a-b}{\sqrt{1 + 0} + \sqrt{1 + 0}} \\ \\ & = \frac{a-b}{2} \\ \end{split}$$

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Before engaging in all the formal limit-taking, it is worth getting an intuitive grip on what is going on. Here is how I would get to the result, and be sure of the result, before going back, being rigorous, and proving it.

First of all, because squares are involved and the middle term in a square is divisible by 2, I will look at $\sqrt{x^2+2ax}-\sqrt{x^2+2bx}$ instead of the stated values.

Second, because $x$ gets larger without limit, any term not involving $x$ will get forever smaller compared to it. Thus the error from replacing $x^2+2ax$ with $x^2+2ax+a^2$ will get smaller and smaller relative to $x^2+2ax$. It tends to negligibility as $x$ tends to infinity.

So let's do the replacement, and consider $\sqrt{x^2+2ax+a^2}-\sqrt{x^2+2bx+b^2}$. This precisely equals $(x+a)-(x+b)$, which equals $a$-$b$.

Translating back to the language of the original question, I therefore know that the answer ought to be $\frac{1}{2}(a-b)$. In that knowledge, I can go back and follow the same route rigorously, already knowing what the answer should be and already understanding the steps to be taken to reach it.

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