Proof that square root of a sequence converges to the square root of the limit of the sequence.

Question

I've seen other posts around here regarding this issue as well, so I apologise for the repetition. But I need help regarding a specific part of the proof that I don't see covered elsewhere. I was doing a question and it had some guiding sub-questions which I completed, but I cannot see how these sub-questions contribute to the final part of the proof using the epsilon-delta definition of the limit of sequences. Here's what I was asked to show in the sub-questions, which I managed to: $$ If \ lim_{\to\infty}a_n=L,then\ when \ n>N_1:N_1\in\mathbb{Z}^+, a_n<\frac{L}{4} \ and \\ |\sqrt{a_n}-\sqrt{L}|<\frac{2|a_n-L|}{3\sqrt{L}} $$ Now, I want to show that $|\sqrt{a_n}-\sqrt{L}|<\epsilon$ using these 2 inequalities. I tried $$ |\sqrt{a_n}-\sqrt{L}|<\frac{2|a_n-L|}{3\sqrt{L}}<\frac{2}{3\sqrt{L}}*\frac{3L}{4}=\frac{\sqrt{L}}{2}<\frac{3L}{4}=\epsilon $$ But I know this is wrong because this epsilon is the one I selected for the sequence ${a_n}$ to get the first inequality, and I must prove $\forall\epsilon$. How do the 2 inequalities apply to the proof, and do I need to still use the same $N_1$ for the proof next, or do I have another $N_2$ to create another special epsilon? Thank you very much for your help.

Answer 1

I suppose your $L $ is positive.

You did not make full use of the fact that $a_n \to L$ and that is why you got stuck. Choose $N_1$ such that $n >N_1$ implies $|a_n-L| <\epsilon \sqrt L$. Then you get $|\sqrt a_n -\sqrt L|=\frac {|a_n-L|} {\sqrt a_n+\sqrt L}\leq \frac {|a_n-L|} {\sqrt L}<\epsilon$ for $n >N_1$.

The result is also true for $L=0$ and the proof is much simpler in that case.