Proving the Limit of sqrt(a(n)^2) = a^2

[LMKIYHAQ]

Homework Statement

I am trying to prove that given a_n\geq0 for all n , and
lim(a_n=a), that lim(sqrt(a(n)^2))=a^2.

Homework Equations

The Attempt at a Solution

I have multiplied the bottom and top by the conjugate but I cannot find what to set as a lower bound for the absolute value of sqrt(an)+sqrt(a).

Please help.

 

[tiny-tim]

I am trying to prove that given a_n\geq0 for all n , and
lim(a_n=a), that lim(sqrt(a(n)^2))=a^2.
…
I have multiplied the bottom and top by the conjugate but I cannot find what to set as a lower bound for the absolute value of sqrt(an)+sqrt(a).

Hi LMKIYHAQ!

I don't get it … for a_n ≥ 0, sqrt(a(n)^2) = a_n, and so lim(sqrt(a(n)^2)) = a, doesn't it?

 

[statdad]

I agree with tiny-tim: whether you write it as

<br /> \lim_{n \to \infty} \sqrt{a_n^2}<br />

or

<br /> \lim_{n \to \infty} \sqrt{a_n}^2<br />

the limit certainly is not a^2.

 

[Unassuming]

You mean

 

[LMKIYHAQ]

Darn. I wrote the problem down wrong everybody. I am sorry, I hope you are all good enough s.t. you didn't spend much time thinking about that! I am sorry.

How do I delete a thread?

P.S. i meant lim(sqrt(a(n)))=a(n), and I figured it out.