Analysis: Sequence convergence with Square Roots

[clifsportland]

Homework Statement

Given Lim Cn=c, Prove that Lim$$\sqrt{Cn}$$=$$\sqrt{c}$$

Homework Equations

We are working from the formal definition: for all $$\epsilon$$, there exists an index N such that For all n>=N, |Cn-c|<$$\epsilon$$

The Attempt at a Solution

We as a group have attempted this several times from several directions, too many to post here. we've been working for over an hour with no results. Please HElp!

 

[Mathdope]

So you can't use any limit theorems?

 

[Rainbow Child]

Let $\epsilon_1>0\Rightarrow \epsilon_1\,(\epsilon_1+2\sqrt{c})>0$ then

$$\exists \,N:|C_n-c|<\epsilon_1\,(\epsilon_1+2\sqrt{c}), \forall n\geq N$$

Thus $\forall\, n\geq N$

$$C_n-c<\epsilon_1\,(\epsilon_1+2\sqrt{c}) \Rightarrow C_n<\epsilon_1^2+2\,\epsilon_1\,\sqrt{c}+c\Rightarrow C_n<(\epsilon_1+\sqrt{c})^2\Rightarrow \sqrt{C_n}<\epsilon_1+\sqrt{c}$$

Now choose $\epsilon=max\Big(\epsilon_1\,(\epsilon_1+2\sqrt{c}),\epsilon_1+2\sqrt{c}\Big)$ and apply the triangle inequality at $|\sqrt{C_n}-\sqrt{c}|$.

 

[matness]

[itex] |c_n -c | = | \sqrt{c_n} - \sqrt{c}| |\sqrt{c_n} +\sqrt{c}| [/tex] should work i think . you can prove [tex]|\sqrt{c_n} +\sqrt{c}| [/itex] is bounded using the definition you gave in 2 and choose a new epsilon
Of course i am assuming we are talking about a nonegative sequence