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I am having some trouble showing $ \lim_{ x \rightarrow 4} \frac{1}{\sqrt{x}} = \frac{1}{2}. $

I chose $\delta$ to be the minimum of $(\frac{2}{1+2\epsilon})^2 - 4$ and $ 4 - (\frac{2}{1-2\epsilon})^2$. It works and was inspired by a picture, but I feel there is an easier trick.

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Given $\varepsilon>0$, choose $\delta=\frac{4}{(2\varepsilon +1)^2}$. Then, if |x-4|<\delta, then $\left|\frac{1}{\sqrt{x}}-\frac 12\right|$ = \left|\frac{2-\sqrt{x}}{2\sqrt{x}}\right|<\varepsilon holds because 2<\sqrt{x}(2\varepsilon+1). You get this last inequality via the triangle inequality.

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    @nbubis: Indeed, I found a delta for x->16, not x->4, but the reasoning is the same. I'll edit out my answer.2012-04-22