Why does $\lim_{x\to 0+} \sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}}$ equal zero?

Question

Limit $$\lim_{x\to 0+} \sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}}$$ equals zero.

Could you help me prove it?

Answer 1

Let $y = \dfrac1x$. We then have $$L = \lim_{x \to 0^+} \left(\sqrt{\dfrac1x+2} - \sqrt{\dfrac1x}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) \times \dfrac{\sqrt{y+2} + \sqrt{y}}{\sqrt{y+2} + \sqrt{y}}$$ This gives us $$L = \lim_{y \to \infty} \dfrac{y+2 - y}{\sqrt{y+2} + \sqrt{y}} = \lim_{y \to \infty} \dfrac2{\sqrt{y+2} + \sqrt{y}}$$ Can you now finish it off?

Answer 2

Hint: $$\sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}} = \dfrac{2}{\sqrt{\dfrac {1}{x}+2}+\sqrt{\dfrac {1}{x}} }.$$

Answer 3

$$\sqrt{\frac1x+2}-\sqrt\frac1x=\frac2{\sqrt{\frac1x+2}+\sqrt\frac1x}=\frac{2\sqrt x}{\sqrt{2x+1}+1}\xrightarrow[x\to 0]{}\frac 0{1+1}=0$$