How to find limits at infinity algebraically for $f(x)=\sqrt{x+2}-\sqrt{x}$?

Question

For example the function $f(x)=\sqrt{x+2}-\sqrt{x}$ How do I find the limit as the function approaches infinity without graphing? Thanks a lot in advance!

Answer 1

$$\begin{array}{rcl} \sqrt{x+2}-\sqrt x &=& \dfrac{(\sqrt{x+2})^2 - (\sqrt x)^2}{\sqrt{x+2} + \sqrt x} \\ &=& \dfrac{2}{\underbrace{\sqrt{x+2}}_\infty+\underbrace{\sqrt x}_\infty} \\ &\to& 0 \end{array}$$

Answer 2

$$ f(x)=\sqrt{x+2} - \sqrt{x} = \sqrt{x} \left(\sqrt{x+2\over x} - 1 \right) = \sqrt{x} \left(\sqrt{1+{2\over x}} - 1 \right) $$ $$ =\sqrt{x} \left(1+{1\over x}+O\Big({1\over x^2}\Big) - 1 \right) ={1\over\sqrt{x}} + O\Big({1\over x^{3/2}}\Big) \ \to \ 0 \quad\mbox{ as }x\to\infty. $$

Answer 3

By the mean value theorem, $f(x+c)-f(x) =cf'(z) $ where $x \le z \le x+c$.

Therefore, if $f'(x) \to 0$ as $x \to \infty$, $f(x+c)-f(x) \to 0$ for any fixed $c$.

If $f(x) = \sqrt{x}$, $f'(x) =\dfrac1{2\sqrt{x}} \to 0 $ as $x \to \infty$. Therefore $\sqrt{x+c}-\sqrt{x} \to 0 $ as $c \to \infty$.

Note that the MVT gives the bounds $\dfrac{c}{2\sqrt{x}} \ge \sqrt{x+c}-\sqrt{x} \ge \dfrac{c}{2\sqrt{x+c}} $.