epsilon-delta proof Would anyone please give me a epsilon-delta proof:
when $x$ approaches to infinity, $$x\sqrt{x+2}-\sqrt x=\infty\;.$$ 
What I did was: 
$$\left(x\sqrt{x+2}-\sqrt x\right)\cdot\frac{\sqrt{x+2}+\sqrt x}{\sqrt{x+2}+\sqrt x}
=\frac{2x}{\sqrt{x+2}+\sqrt x}$$
Then, what is the next step? 
 A: I would take a cruder approach. Let $x$ be positive. Then $\sqrt{x+2}\gt \sqrt{x}$.
It follows that
$$x\sqrt{x+2}-\sqrt{x}\gt x\sqrt{x}-\sqrt{x}=(x-1)\sqrt{x}.$$
Furthermore, if $x\gt 1$, then $(x-1)\sqrt{x}\gt x-1$.
Now it should be easy, given any $K$, however large, to come up with an $L$ such that if $x\gt L$, then $x-1 \gt K$.
Remark: We have not been asked, given $K$, to come up with the cheapest $L$ such that for $x$ beyond $L$, we have $x\sqrt{x+2}-\sqrt{x} \gt K$.
A: In response to edit, when you tried to multiply the equation by the conjugate you made a mistake $\sqrt{x+2} + \sqrt{x}$ is conjugate to $\sqrt{x+2}-\sqrt{x}$, not $x\sqrt{x+2} - \sqrt{x}$. Instead what you can do is factor our $x$ first
$$x\sqrt{x+2} - \sqrt{x} = x\left(\sqrt{x+2} - \sqrt{\frac{1}{x}}\right)$$
I would then attempt to prove that the latter expression in brackets approaches infinity.
A: Also, are you sure this is the right question? The two terms are so different in value
($O(x^{3/2})$ and $\sqrt{x}$) that this is an unlikely question.
Perhaps the real question is to show that 
$\lim_{x\to\infty}(x\sqrt{x+2}-x \sqrt x)=\infty$.
Then the conjugate expression would be of use.
