# Homework Help: Proving a limit theorem of a sequence (square root)

1. Sep 29, 2009

### shoescreen

1. The problem statement, all variables and given/known data
given the sequence {a_n} converges to A (non zero), show sqrt(a_n) = sqrt(A)

2. Relevant equations

3. The attempt at a solution
I've tried to expand |sqrt(a_n) - sqrt(A)| as |a_n - A|\|sqrt(a_n) + sqrt(A)| since that gives me the numerator to work with, but I can't figure out how to work with sqrt (a_n).

I'm trying a standard n* epsilon proof, that is finding an n* such that the above quantity is less than any positive real number epsilon.

Thanks for the help!

Last edited: Sep 29, 2009
2. Sep 30, 2009

### fmam3

Suppose $$\lim a_n = A$$ and you want to show $$\lim \sqrt{a_n} = \sqrt{A}$$.

So, right now you need a bound for the denominator. Assuming that we are working with the reals (i.e. no complex numbers for this problem), then see that for the square root to be defined, we must have that $$a_n, A \geq 0$$. Thus, that means we can simplify by writing $$| \sqrt{a_n} + \sqrt{A} | = \sqrt{a_n} + \sqrt{A} \geq \sqrt{A}$$ right?

That should be enough for you to finish the proof...

3. Sep 30, 2009

### shoescreen

hmmm, but the inequality is in the wrong direction. If want to compare |a_n - A|\|sqrt(a_n) + sqrt(A)| with a smaller sequence, I need something bigger in the denominator, and since root(A) <_ root (a_n) + root(A), I can't guarantee anything about the original sequence.

4. Sep 30, 2009

### fmam3

Note that WLOG, we can assume that $$A \ne 0$$ since if $$A = 0$$, then the result is trivial.

In fact, we can even further say that we can assume, WLOG that $$A > 0$$ (note the strict inequality). Then, if we have $$\sqrt{a_n} + \sqrt{A} > A$$, then does that not imply $$\frac{1}{\sqrt{a_n} + \sqrt{A}} < \frac{1}{\sqrt{A}}$$? Can you finish the rest of the proof from here?

5. Sep 30, 2009

### shoescreen

I take back what i said before, the inequality is in the correct direction. It's frustrating, I had all of this before yet trashed it all because of that inequality sign. Ugh, guess i wasn't in math mode.

Thanks for the help!

6. Sep 30, 2009

No prob :)

7. Feb 3, 2012

### skoomafiend

bumping an old post, but i am having trouble with a similar problem.

how would one continue from where he ended off?

thanks!

8. Feb 3, 2012

### jbunniii

Using the equality obtained in the excerpt you quoted, you would write

$$|\sqrt{a_n} - \sqrt{A}| = \frac{|a_n - A|}{|\sqrt{a_n} + \sqrt{A}|} < \frac{|a_n - A|}{\sqrt{A}}$$

Now, can you make the numerator as small as you like by choosing a large enough n? If so, what can you conclude?

9. Feb 3, 2012

### skoomafiend

this is about where i'm getting stuck.
would be able to rephrase that question in any other way?

10. Feb 3, 2012

### jbunniii

You're trying to show that

$$\lim_{n \rightarrow \infty} \sqrt{a_n} = \sqrt{A}$$

What does this statement mean? What is the definition of the limit of a sequence?

11. Feb 3, 2012

### skoomafiend

i am trying to show that for all ε > 0, there is an N such that for all n>=N i'll have,

$$|\sqrt{a_n} - \sqrt{A}| < ε$$

i'm not understanding how to relate this (below) to my epsilon.

$$|\sqrt{a_n} - \sqrt{A}| < \frac{|a_n - A|}{\sqrt{A}}$$

12. Feb 3, 2012

### jbunniii

You know that

$$\lim_{n \rightarrow \infty} a_n = A$$

So given any $\epsilon' > 0$ (I intentionally put a ' on the $\epsilon$ to indicate that it's a different variable) there is an N such that

$$|a_n - A| < \epsilon'$$

for all $n \geq N$.

Now think about what choice of $\epsilon'$ would be useful to get the result you require.

13. Feb 3, 2012

### skoomafiend

$$ε' = ε \sqrt{A}$$

would this work?

thank you for all your help so far.

14. Feb 3, 2012

### jbunniii

Looks promising. Why don't you write out your argument in detail and we'll see if it works.

15. Feb 3, 2012

### skoomafiend

Since,
$$|a_n - A| < ε$$

We have that for,
$$ε' > 0$$
there exists N such that for all n>=N,

$$|\sqrt{a_n} - \sqrt{A}| < ε'$$

so,

$$|\sqrt{a_n} - \sqrt{A}| < ... < \frac{|a_n - A|}{\sqrt{A}} < \frac {ε'}{\sqrt{A}} = ε$$

where
$$ε' = ε\sqrt{A}$$

I feel as though i'm missing some critical parts of the argument.

16. Feb 3, 2012

### jbunniii

You have the right idea, but you didn't arrange the proof quite correctly.

Try starting as follows:

Let $\epsilon > 0$. We seek an $N$ such that whenver $n \geq N$, the following inequality is satisfied:

$$|\sqrt{a_n} - \sqrt{A}| < \epsilon$$

We know that

$$\lim_{n \rightarrow \infty} a_n = A$$

Therefore, given $\epsilon' > 0$, ...

17. Feb 3, 2012

### skoomafiend

i'm definitely having trouble arranging the proof correctly.
the given ϵ′ is for $$| a_n - A| < ϵ'$$ ?

18. Feb 3, 2012

### jbunniii

Right. Given $\epsilon' > 0$, there exists an $N'$ such that

$$|a_n - A| < \epsilon'$$

whenever $n > N'$.

I can choose $\epsilon'$ to be whatever I like, and there is guaranteed to be a corresponding $N'$ that makes the above inequality true. So I choose $\epsilon' = \epsilon\sqrt{A}$...

19. Feb 3, 2012

### skoomafiend

now do i say that for every ϵ > 0, there exists an N such that all n >= N where $$|\sqrt{a_n} - \sqrt{A}| < ϵ = \frac{ϵ'}{\sqrt{2}}$$

20. Feb 3, 2012

### jbunniii

Now use the chain of inequalities from message #8.