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I've been struggling to put together limit proofs that hold up, thanks to some previous questions I learned a few things. I hope that this time, I've managed to put together a good proof:

Given $\epsilon > 0, \ \exists N \ s.t. \ \forall n \geq N$ so that:

$|x_n - 4| < \epsilon$

There exists $N_1 \ s.t. \forall n \geq N_1$:

$|x_n - 4| < 1$

$3 < x_n < 5$

$\sqrt{3} < \sqrt{x_n} < \sqrt{5}$ and $\sqrt{x_n}$ is shown to be bounded below by $\sqrt{3}$ for $N_1$

Let $N_* = \max(N_1, N)$, then $\forall n \geq N_*$<

$|x_n - 4| = |\sqrt{x_n} - 2||\sqrt{x_n} + 2| < \epsilon$

$\epsilon >|\sqrt{x_n} - 2||\sqrt{x_n} + 2| > \sqrt{3}|\sqrt{x_n} - 2|$ and thus

$\sqrt{3}|\sqrt{x_n} - 2| < \epsilon, \ \ |\sqrt{x_n} - 2| < \frac{\epsilon}{\sqrt{3}}$


Much simpler proof given in the comments!

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  • $\begingroup$ Do you have alrealdy learnt that if $f$ is a continuous function and $x_n \rightarrow a$ then $f(x_n) \rightarrow f(a)$? If your answer is yes, you just need to show that $x \rightarrow \sqrt(x)$ is continuous in $(0, +\infty)$. $\endgroup$ – Luísa Borsato Jan 27 '17 at 14:45
  • $\begingroup$ Ah, yes sorry! I should have clarified, it needs to be an $\epsilon$ proof. $\endgroup$ – student_t Jan 27 '17 at 14:51
  • $\begingroup$ Nice question, and your efforts in tackling the problemmake the question much easier for people to answer. Well done +1 $\endgroup$ – Kevin Jan 27 '17 at 15:16
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In the second-to-last line it looks like you should have $(\sqrt 3 + 2)|\sqrt{x_n} - 2|$ rather than just $\sqrt 3 |\sqrt{x_n} - 2|$.

I would suggest a simplification, however. You observed that $|\sqrt{x_n} - 2||\sqrt{x_n} + 2| = |x_n - 4|$ and this leads right away to $$|\sqrt{x_n} - 2| \le \frac{|x_n - 4|}{2}.$$

So $|x_n - 4| < \epsilon $ implies $|\sqrt{x_n} - 2| < \epsilon$.

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  • $\begingroup$ Yes! Of course, thank you! $\endgroup$ – student_t Jan 27 '17 at 14:55
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I think the required proof can have a simpler look.

We suppose that $x_n\ge0$ for all $n$ and that $\lim_{n\to\infty}x_n=4$.

For all $n$, we have :

$$\left|\sqrt{x_n}-2\right|=\frac{\left|x_n-4\right|}{\sqrt{x_n}+2}\le\frac{\left|x_n-4\right|}{2}$$

Consider $\epsilon > 0$.

We know that there exists $N\in\mathbb{N}$ such that $n\ge N\implies\left|x_n-4\right|\le2\epsilon$

Hence, $n\ge N\implies\left|\sqrt{x_n}-2\right|\le\epsilon$

as required.

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  • $\begingroup$ Haha yes this is much simpler, thank you, I clearly overcomplicated things. $\endgroup$ – student_t Jan 27 '17 at 14:56

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