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

Question

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?

Answer 1

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

Answer 2

Your answer is correct. You could also use binomial series:

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

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

Answer 3

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 $\dfrac{1}{2}$.