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$∀n ≥ 1 : a_n ≥ 0$, and limit $\lim_{n→∞} a_n = L$. Prove that $[\lim_{n→∞} √a_n] = √L$.

For what I did is,

$$\lim_{n→∞} a_n = [\lim_{n→∞} √a_n ][\lim_{n→∞} √a_n]$$

$$[\lim_{n→∞} √a_n ][\lim_{n→∞} √a_n] = L$$

$$[\lim_{n→∞} √a_n]² = L$$

$$[\lim_{n→∞} √a_n] = √L$$

but i dont how the prove $\lim_{n→∞} √a_n$ exists

hint from my professor {show $L ≥ 0$, separate the cases $L > 0$ and $L = 0$, make use of $|√a_n - √L|·(√a_n + √L) = |a_n - L|$ }

Please help!! Thank you!!

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The limit exists because $\sqrt{\space}$ is a continuous function and because $\displaystyle \lim \bigl( (a_n)_{n\in\mathbb{N}}\bigr)$ exists. –  Git Gud Jan 29 '13 at 21:33
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2 Answers 2

up vote 0 down vote accepted

I think you can interpret that hint in the following way:

If $L=0$, then: $∀ε>0: ∃N∈ℕ: ∀n∈ℕ, n>N: |a_n| < ε^2$, and therefore $|√a_n| < ε$, this shows convergence.

If $L>0$, then: $∀ε>0: ∃N∈ℕ: ∀n∈ℕ, n>N: |a_n - L| < ε√L$ and therefore $|√a_n - √L| = \tfrac{|a_n - L|}{√a_n + √L} ≤ \tfrac{|a_n - L|}{√L} < ε$, which shows convergence.

Note, that you can really choose an $N$ to go below $ε^2$ and $ε√L$ for an arbitrary $ε>0$ because they too are arbitrary positive real numbers.

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is it because they are both positive real number, then i can use what i did to follow your theorem?? –  Paul Jan 30 '13 at 1:11
    
I don't understand your question. –  k.stm Jan 30 '13 at 8:41
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This is equivalent to say that $x\mapsto \sqrt x$ is a continuous function.

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but may i know how to make use of the hint of my professor in order to get some marks? –  Paul Jan 29 '13 at 21:36
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