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Using epsilon and delta proof this :

$$\lim_{x \to 3} \sqrt{x+1}\neq1$$

Exist $\varepsilon>0$ All $\delta>0$ so for some $x$ that appiles $0<|x-3|<\delta$ and however $|\sqrt{x+1}-1|\geq\varepsilon$

I don't really know how to approach this, any help will be appreciated.

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  • $\begingroup$ The limit definitely exists, it's just not equal to 1. $\endgroup$ – Tavian Barnes Aug 13 '14 at 16:54
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Choose any $\varepsilon < 1$. For whatever $\delta$ you can choose $x = \min\lbrace 4, 3+\frac{\delta}{2}\rbrace$ So:

$$\vert \sqrt{x+1}-1\vert > \vert \sqrt{3+1}-1\vert = 1 > \varepsilon$$

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    $\begingroup$ Can you explain did you manage to conclude $x$=min{4,3+$\frac{\delta}{2}$}, Thanks $\endgroup$ – JaVaPG Aug 13 '14 at 16:36
  • $\begingroup$ @JaVaPG There are many values of $x$ that would do the trick, you could simply choose $x = 3+\delta/2$ Or any value in $(3,3+\delta)$ for that matter. The inequalities bellow would still work. Choosing $x = \min\lbrace 4, 3+\frac{\delta}{2}\rbrace$ was the first thing that popped into my mind so I went with it. If it made things more complicated for you, just take $x = \frac{\delta}{2}$. $\endgroup$ – Darth Geek Aug 13 '14 at 16:41

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