How to Solve $\lim_{x\to \infty}\sqrt{x^2 + ax} - \sqrt{x^2 + bx}$

Question

I can't quite figure out how to manipulate this into a determinate form — should I try rationalizing it by multiplying by the conjugate, completing the square, or something like that?

Note: I'm in precalc, so haven't learned any fancy calculus theorems, etc.

Related: $\lim\limits_{x \to \infty }\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)!$, $\lim\limits_{x \to \infty }\left(\sqrt{x^2+3x}-\sqrt{x^2+x}\right)!$, I'm sure you can extend that technique for the general case with $a$ and $b$. — Workaholic, May 19 '16 at 20:24
Hint: $\frac{\sqrt{x^2+ax}+\sqrt{x^2+bx}}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$. — vadim123, May 19 '16 at 20:25
You may also 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 standard techniques for finding limits of the form, $$\lim_{x\to\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.$$ See in particular this answer by @Sasha, which should be at your level, assuming you know how to handle the $\Sigma$-notation. — Workaholic, May 19 '16 at 20:33
Actually this turns out to be a duplicate, as I initially guessed, of this one. — Workaholic, May 19 '16 at 20:36

score 5 · Accepted Answer · edited May 19 '16 at 20:38

5

$$ \begin{aligned} \sqrt{x^{2}+ax}-\sqrt{x^{2}+bx}&=\frac{\sqrt{x^{2}+ax}-\sqrt{x^{2}+bx}}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+bx}}\cdot(\sqrt{x^{2}+ax}+\sqrt{x^{2}+bx})\\[6pt] &=\frac{x^{2}+ax-x^{2}-bx}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+bx}}\\[6pt] &=\frac{(a-b)x}{\sqrt{x^{2}+ax}+\sqrt{x^{2}+bx}}\\[6pt] &=\frac{a-b}{\sqrt{1+a/x}+\sqrt{1+b/x}}. \end{aligned} $$

Now take the limit.

edited May 19 '16 at 20:38

Michael Hardy

1

answered May 19 '16 at 20:27

ervx

12,208

score 2 · Answer 2 · answered May 19 '16 at 20:28

2

Hint:

Here's an overview of what you should do:

Rationalize by multiplying by the conjugate.
Simplify the numerator.
Factor an $x$ out of the denominator and reduce the whole fraction.
Take the limit of the new expression.

answered May 19 '16 at 20:28

Noble Mushtak

18,402
28
44

How to Solve $\lim_{x\to \infty}\sqrt{x^2 + ax} - \sqrt{x^2 + bx}$

2 Answers2