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Prove: $s_n \to s \implies \sqrt{s_n} \to \sqrt{s}$ by the definition of the limit. $s \geq 0$ and $s_n$ is a sequence of non-negative real numbers.

This is my preliminary computation:

$|\sqrt{s_n} - \sqrt{s}| < \epsilon$

multiply by the conjugate:

$|\dfrac{s_n - s}{\sqrt{s_n}+\sqrt{s}}| < \epsilon$

Thus we can use the fact that $|\sqrt{s_n} - \sqrt{s}| < \dfrac{|s_n - s|}{\sqrt{s}} < \epsilon$

After this I am lost...

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    It is formally written ;) . You've proven in precalculus courses that $\bigl||a|-|b|\bigr| \le |a| + |b|$ right? What happens with the absolute value of a positive number?2012-10-03

4 Answers 4

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If both $s$ and $s_n$ are non-negative

$ |\sqrt{s}-\sqrt{s_n}|^2 \le |\sqrt{s}-\sqrt{s_n}||\sqrt{s} + \sqrt{s_n}|. $

Step by Step :)

Since both $s$ and $s_n$ are non-negative

$ |\sqrt{s}-\sqrt{s_n}| \le |\sqrt{s} + \sqrt{s_n}| $

this is clear because the result of substracting a non-negative number from another is always less than the result of adding it, then

$ |\sqrt{s}-\sqrt{s_n}|^2 \le |\sqrt{s}-\sqrt{s_n}| \cdot |\sqrt{s}+\sqrt{s_n}| = |s - s_n| $

and you are done!

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    @CodeKingPlusPlus No, what you need is to prove that |s - s_n| < \epsilon implies that _For all_ \epsilon_1 > 0 _there exist an_ $N$ _such that_ |\sqrt{s} - \sqrt{s_n}| < \epsilon_1 _for_ n > N. Now, you've proven that such $N$ exist when $\epsilon_1 = \sqrt{\epsilon}$2012-10-03
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ADD You got to

$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }}$

Since $s_n\to s$, for every $\epsilon >0$ there is an $n_0$ for wich $\left| {{s_n} - s} \right| < \varepsilon \sqrt s $ whenever $n\geq n_0$ (i.e. $\varepsilon \sqrt s$ is also an $\epsilon'>0$). Then, for this $n_0$, $\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }} < \frac{{\varepsilon \sqrt s }}{{\sqrt s }} = \varepsilon $

which means $\sqrt {s_n}\to\sqrt s$.


You're almost done. You arrived at

$\left|\dfrac{s_n - s}{\sqrt{s_n}+\sqrt{s}}\right| $

You know that $s_n\to s$, so you can make $|s_n-s|$ as small as you wish. Now, we need to know how to handle $\sqrt{s_n}+\sqrt{s}$. Since $s_n\to s$, there is an $n_0$ for wich

$|s-s_n|<3s/4$

Since $s_n>0$,$s\geq 0$, then

$s-s_n\leq|s-s_n|<3s/4$

This means that

$s_n>s/4$

then

$2\sqrt {{s_n}} > \sqrt s $

or, since $\sqrt s>0$

$\eqalign{ & 2\sqrt {{s_n}} + 2\sqrt s > 3\sqrt s \cr & \sqrt {{s_n}} + \sqrt s > \frac{{3\sqrt s }}{2} \cr & \frac{1}{{\sqrt {{s_n}} + \sqrt s }} < \frac{2}{{3\sqrt s }} \cr} $

Again, since $s_n\to s$, there is an $n_1$ for wich

$|s-s_n|<\epsilon \frac{3\sqrt s}{{2 }}$

Then, taking $n\geq \max\{n_0,n_1\}$ we have

$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{2}{{3\sqrt s }}\left| {{s_n} - s} \right| < \frac{2}{{3\sqrt s }}\frac{{3\sqrt s }}{2}\varepsilon = \varepsilon $

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Since $s_n\to s$, for any $\epsilon$, we can find an $N$ so that for all $n\ge N$ we have $|s_n-s|<\epsilon\sqrt{s}$. Then $ |\sqrt{s_n}-\sqrt{s}|=\left|\frac{s_n-s}{\sqrt{s_n}+\sqrt{s}}\right|\le\left|\frac{s_n-s}{\sqrt{s}}\right| $ to get that $ |\sqrt{s_n}-\sqrt{s}|\le\epsilon $

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well, you can actually take $\epsilon={\sqrt{s}}\epsilon'$ for any $\epsilon'>0$ given that for all $n>N, \epsilon'>|s_n-s|$

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    It's the other way round: you're given $\epsilon$, then take $\epsilon'= \epsilon/\sqrt s$ and then an $N$ for $\epsilon'$.2012-10-03