Prove $\lim_{x\to\infty} \left( \sqrt{x+1} - \sqrt{x} \right) = 0$ 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.   
 A: 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.
A: 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.
A: 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
A: $$
0 < \sqrt{x+1} - \sqrt{x} = \int_x^{x+1} \frac1{2\sqrt{t}}dt \le \int_x^{x+1} \frac1{2\sqrt{x}}dt = \frac1{2\sqrt{x}} \rightarrow 0
$$
We have $\frac1{2\sqrt{t}} \le \frac1{2\sqrt{x}}$ because $\frac1{2\sqrt{t}}$ is a decreasing function on the interval.
A: $(x+\frac1{2x})^2
=x^2+1+\frac1{4x^2}
\gt x^2+1
$
so
$x+\frac1{2x}
\gt \sqrt{x^2+1}
$
so
$\sqrt{x^2+1}-x
\lt \frac1{2x}
$
so
$\sqrt{x+1}-\sqrt{x}
\lt \frac1{2\sqrt{x}}
$.
