# Limit of a sum of two oddly specific square roots at infinity

I can't wrap my head around this one. How would you approach finding a limit like this:

$$\lim_{x\to\infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-2x}\right)$$

I managed to simplify a bit:

$$\lim_{x\to\infty}\left(\sqrt{x}\cdot(\sqrt{x+2}-\sqrt{x-2})\right)$$

but it's not far. The similarity between the roots must be the key, but I've no idea where to start. I welcome any pointers, and appreciate even more if there is a general solution. But I fear this is a specific case.

# EDIT:

WolframAlpha gave an answer: 2, but I'm still interested in the solution behind it.

• should be a + on the bottom @mfl – John Lou Nov 8 '17 at 13:57

$$\lim_{x\to\infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-2x}\right)=\lim_{x\to\infty}\frac{\left(\sqrt{x^2+2x}-\sqrt{x^2-2x}\right)\left(\sqrt{x^2+2x}+\sqrt{x^2-2x}\right)}{\left(\sqrt{x^2+2x}+\sqrt{x^2-2x}\right)}=$$ $$\lim_{x\to\infty}\frac{x^2+2x-(x^2-2x)}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}=\lim_{x\to\infty}\frac{4x}{|x|\sqrt{1+\frac{2}{x}}+|x|\sqrt{1-\frac{2}{x}}}=$$ $$=\lim_{x\to+\infty}\frac{4x}{|x|\left(\sqrt{1+\frac{2}{x}}+\sqrt{1-\frac{2}{x}}\right)}=\lim_{x\to+\infty}\frac{4}{\left(\sqrt{1+\frac{2}{x}}+\sqrt{1-\frac{2}{x}}\right)}=\frac{4}{2}=2$$

There's a standard trick for limits that involve sums of differences of square roots: "Multiplication by the conjugate."

In this case, applying the trick means that we rewrite the argument of the limit as $$\left(\sqrt{x^2 + 2 x} - \sqrt{x^2 - 2 x}\right) \cdot \frac{\sqrt{x^2 + 2 x} + \sqrt{x^2 - 2 x}}{\sqrt{x^2 + 2 x} + \sqrt{x^2 - 2 x}}.$$ What happens when we distribute and simplify?

\begin{align} \lim_{x \to \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-2x}\right) &= \lim_{x \to \infty}\left(\sqrt{x^2+2x}-\sqrt{x^2-2x}\right) \cdot \frac{\sqrt{x^2+2x}+\sqrt{x^2-2x}}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\\ &=\frac{x^2+2x-x^2+2x}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\\ &=\frac{4x}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\\ &=\frac{4\sqrt x}{\sqrt{x+2}+\sqrt{x-2}}\\ &\approx \frac{4\sqrt x}{\sqrt{x}+\sqrt{x}}\\ \lim_{x \to \infty} \frac{4\sqrt x}{\sqrt{x+2}+\sqrt{x-2}}&= 2 \end{align}

• Is that sort of approximation standard practice? Seems odd. Maybe justified given the infinity isn't affected much by the constants.. – Felix Nov 8 '17 at 14:03
• The constants make no difference to the function as it approaches infinity. If you prefer, you can do what @raffaele did and divide by $x^2$ instead of $\sqrt x$. – John Lou Nov 8 '17 at 14:05

Here is a method that would also work with a cubic root

Let $f \left(t\right) = \sqrt{1+t}$ and

$${\varepsilon} \left(t\right) = \frac{f \left(t\right)-f \left(0\right)}{t-0}-{f'} \left(0\right)$$

It is well known that $\lim_\limits{t \rightarrow 0} {\varepsilon} \left(t\right) = 0$. Using $f \left(0\right) = 1$ and ${f'} \left(0\right) = \frac{1}{2}$, we deduce

$$\sqrt{1+t} = f \left(t\right) = 1+\frac{1}{2} t+t {\varepsilon} \left(t\right)$$

It follows that when $x > 0$

$$\sqrt{{x}^{2}+2 x} = \sqrt{{x}^{2}} \sqrt{1+\frac{2}{x}} = x\ f\left(\frac{2}{x}\right)= x \left(1+\frac{1}{x}+\frac{2}{x} {\varepsilon} \left(\frac{2}{x}\right)\right) = x+1+ 2{\varepsilon} \left(\frac{2}{x}\right)$$

Similarly

$$\sqrt{{x}^{2}-2 x} = x-1-2{\varepsilon} \left(\frac{{-2}}{x}\right)$$

Hence

$${\lim }_\limits{x \rightarrow \infty } \left(\sqrt{{x}^{2}+2 x}-\sqrt{{x}^{2}-2 x}\right) = {\lim }_\limits{x \rightarrow \infty } \left(2+2{\varepsilon} \left(\frac{2}{x}\right)+2{\varepsilon} \left(\frac{{-2}}{x}\right)\right) = 2$$