# Evaluate: $\lim\limits_{x \to \infty} (\sqrt{x+2}-\sqrt{x})$

Evaluate: $$\displaystyle\lim_{x \to \infty} (\sqrt{x+2}-\sqrt{x})$$

I was given this problem. I'm not sure how to tackle it. $$\infty-\infty$$ is in indeterminate form, so I need to get it into a fraction form in order to solve. I did this by pulling an $$x^2$$ out of it: $$\lim_{x\to\infty}x\left(\sqrt{\frac1x+\frac2{x^2}}-\sqrt{\frac1x}\right)\\=\lim_{x\to\infty}\frac{\left(\sqrt{\frac1x+\frac2{x^2}}-\sqrt{\frac1x}\right)}{\frac1x}$$ Now taking the derivative: $$\lim_{x\to\infty}\Large\frac{\frac{\frac{-1}{x^2}+\frac{-4}{x^3}}{2\sqrt{\frac1x+\frac2{x^2}}}-\frac{-\frac1{x^2}}{2\sqrt{\frac1x}}}{-\frac1{x^2}}\\=\lim_{x\to\infty}\frac{-x^2\left(\frac{-1}{x^2}+\frac{1}{x^2}\right)}{2\sqrt{\frac1x+\frac2{x^2}}-2\sqrt{\frac1x}}\\=\lim_{x\to\infty}\frac{-x^2\left(0\right)}{2\sqrt{\frac1x+\frac2{x^2}}-2\sqrt{\frac1x}}\\=\lim_{x\to\infty}0=0$$ Since my numerator becomes zero, my entire fraction becomes zero, making zero the answer.

Is this correct? This was very messy - did I miss the easy way to do it? Is there a better way to do problems of this sort?

• Your computation is NOT OK if the numerator becomes $0$ and the denominator also becomes $0$. Dec 30 '19 at 21:53
• @GEdgar - here, my denominator also becomes zero. Does that mean my method of computation did not work?
– Burt
Dec 30 '19 at 21:54
• You can have a look at: Prove $\lim_{x\to\infty} \left( \sqrt{x+1} - \sqrt{x} \right) = 0$. (Change from $x+1$ to $x+2$ does not make that much difference.) May 15 '20 at 13:11

Hint: $$\sqrt{x+2}-\sqrt{x}=\frac{2}{\sqrt{x+2}+\sqrt{x}}$$

• Wow, that was simple:)
– Burt
Dec 30 '19 at 21:48
• Why is this true?
– Burt
Dec 30 '19 at 21:48
• $(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})=(x+2)-x=2$ Dec 30 '19 at 21:49
• This is the usual trick which is used when we have subtraction of square roots. It reminds of the well known trick of multiplying and dividing by a complex conjugate to get rid of the imaginary part.
– Mark
Dec 30 '19 at 21:55
• Called "rationalize the numerator". Dec 30 '19 at 22:01

$$\lim_{x\to\infty}\left(\sqrt{x+2}-\sqrt x\right)=\lim_{x\to\infty}\sqrt x\left(\sqrt{1+\frac2x}-1\right)$$

$$=\lim_{x\to\infty}\sqrt{x}\left(1+\frac1x+o\left(\frac1x\right)-1\right)=\lim_{x\to\infty}\dfrac{\sqrt x}x=0.$$

• But would he get credit for a correct answer with an incorrect method? Dec 30 '19 at 21:54

We have that $$\sqrt{x+2} - \sqrt{x} = \dfrac{(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})}{\sqrt{x+2}+\sqrt{x}} = \dfrac{2}{\sqrt{x-2}+\sqrt{x}}.$$ Hence $$\lim\limits_{x\to\infty} \sqrt{x+2}-\sqrt{x} = \lim\limits_{x\to\infty}\dfrac{2}{\sqrt{x}(\sqrt{1-\frac{2}{x}}+1)}=0.$$

If you were looking for a nonzero answer, you could have tried something like $$\lim\limits_{x\to\infty} \sqrt{x^2+x}-x$$, which evaluates to $$\dfrac{1}{2}$$.

• So was I totally wrong - my answer seems to really simplify to $\frac00$?
– Burt
Dec 30 '19 at 22:52
• you got the wrong denominator for the third last line. Dec 30 '19 at 23:19
• How did I mess up?
– Burt
Dec 31 '19 at 2:13
• $\dfrac{\frac{-\frac{1}{x^2}-\frac{4}{x^3}}{2\sqrt{\frac{1}{x}+\frac{2}{x^2}}}-\frac{-\frac{1}{x^2}}{2\sqrt{\frac{1}{x}}}}{-\frac{1}{x^2}}\neq \dfrac{-x^2(-\frac{1}{x^2}+\frac{1}{x^2})}{2\sqrt{\frac{1}{x}+\frac{2}{x^2}}-2\sqrt{\frac{1}{x}}}$ Dec 31 '19 at 3:04