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How will I know if limits such as $|a_n-L|\Longleftrightarrow|a_n^2-L^2|$ as $a_n \rightarrow L$ are true?

In this case, I know that $|a_n^2-L^2| \Longrightarrow |a_n-L|$ as $a_n \rightarrow L$ is false because $L$ could be negative or that $a_n = L, -L, L, -L, ...$, but how do I determine if $|a_n-L| \Longrightarrow |a_n^2-L^2|$ is true?

Thank you!

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You're missing "$\to 0$" in a handful of places. – Hans Lundmark Oct 6 '11 at 9:05
Okay, thanks! I will edit the question – Juan de la John Oct 6 '11 at 9:11
up vote 2 down vote accepted

If $a_n-L \to 0$, then $a_n \to L$; hence $a_n^2 - L^2 = (a_n - L)(a_n + L) \to 0 \cdot (L+L) = 0$.

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Oh, that was easy haha thanks! I should have thought about that in the first place. – Juan de la John Oct 6 '11 at 9:11
Following your logic, I tried to solve $a_n+1\rightarrow L+1$, changed the variables for clarity $|x_n-M|\rightarrow 0$ then I substituted and got $|a_n-L|\rightarrow 0$. I also tried it with $2a_n \rightarrow 2L$ where I factored the $2$ out and got $|a_n-L|\rightarrow 0$ again. Did I do it correctly? – Juan de la John Oct 6 '11 at 17:24
@Juan: Uhm... To be honest I don't quite understand what you're trying to prove. – Hans Lundmark Oct 6 '11 at 17:38
My apologies, I was just wondering if $[a_n \rightarrow L] \Longleftrightarrow [a_n + 1 \rightarrow L + 1] \Longleftrightarrow [2a_n \rightarrow 2L]$ is indeed true. – Juan de la John Oct 7 '11 at 1:22

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