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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?

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    Brian, That's funny. When you click "edit" the following message is displayed: "*We welcome all constructive edits, but please make them substantial. Avoid trivial, tiny one-letter edits unless absolutely necessary*."2012-09-18

3 Answers 3

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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 $x\sqrt{x+2}-\sqrt{x} \gt K$.

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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.

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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.