Prove $\lim_{x\to\infty} \left( \sqrt{x+1} - \sqrt{x} \right) = 0$

Question

My attempt: I tried manipulating the formula, but I couldn't do anything useful. I tried to find another function $f(x)$ such that $\lim_{x\to\infty} f(x) = 0$ and $f(x) \geq \sqrt{x+1} - \sqrt{x} $ for all $x$. $f(x) = \frac1x$ fails but I think $f(x) = \frac{1}{\sqrt{x}}$ would work (not sure how to verify this). I'm not sure how to proceed.

Answer 1

Try multiplying your expression by $\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}$ and then simplify the numerator and take the limit of this new expression as $x \to \infty$. What do you get?

Note that this is a common method for solving a problem like this. We say $\sqrt{x + 1} + \sqrt{x}$ is the conjugate of $\sqrt{x + 1} - \sqrt{x}$. The purpose of multiplying by this is that the numerator becomes the factored form of the difference of squares (recall: $(a + b)(a - b)= a^{2} - b^{2}$). This gets rid of the square roots in the numerator and allows us to cancel the $x$'s.

Answer 2

For $x\ge 0$:

$\sqrt{1+x}-\sqrt{x}\le \dfrac{1}{\sqrt{x}}$

if and only if

$\sqrt{1+x}\le\sqrt{x}+\dfrac{1}{\sqrt{x}}$

if and only if

$1+x\le x+2+\dfrac{1}{x}$

if and only if

$0\le 1+\dfrac{1}{x}$

True.

Answer 3

Any differentiable function $f$ satisfying $\lim_{x\to \infty}f'(x)=0$ will also satisfy $\lim_{x\to \infty}(f(x+1)-f(x))=0.$ Proof: MVT