0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ should be a + on the bottom @mfl $\endgroup$ – John Lou Nov 8 '17 at 13:57
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

\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}

$\endgroup$
  • $\begingroup$ Is that sort of approximation standard practice? Seems odd. Maybe justified given the infinity isn't affected much by the constants.. $\endgroup$ – Felix Nov 8 '17 at 14:03
  • $\begingroup$ 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 $\sqrt x$. $\endgroup$ – John Lou Nov 8 '17 at 14:05
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.