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If $\lim_{n \to \infty} |a_n| = 0$, then $\lim_{n \to \infty} a_n = 0$

I read the proof by Squeeze Theorem, and it doesn't seem like the limit can only be $0$. I wonder if I can extend it to

If $\lim_{x \to a} |f(x)| = 0$, then $\lim_{x \to a} f(x) = 0$.

If this holds, it would also seem I can extend the idea to functions in higher dimensions, that is

If $\lim_{(x,y) \to (a,b)} |f(x,y)| = 0$, then $\lim_{(x,y) \to (a,b)} f(x,y) = 0$

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4  
Your "extension" to functions fails. Just consider the function which is +1 when $x$ is rational and -1 when $x$ is irrational. –  Old John Feb 6 '13 at 3:58
    
The case you described works because $+0$ or $-0$ are same and this is not true for any other number other than zero and it wont work. –  Phani Raj Feb 6 '13 at 4:02
    
What do you mean by $L$ and $M$? –  Brad Feb 6 '13 at 4:24
    
The original limits. I edited that out, so no longer applies. –  jip Feb 6 '13 at 4:25
2  
The current edit looks good to me. If the limit of the absolute values is zero, that can only happen if the limit of the values is zero. –  Gerry Myerson Feb 6 '13 at 4:26

1 Answer 1

up vote 1 down vote accepted

Assuming $\lim_{n\to\infty} |a_n| = 0$, this means that given any $\varepsilon > 0$, there is an integer $N$ such that for $n \ge N$, $$ \big| |a_n| - 0 \big| < \varepsilon \quad\Leftrightarrow\quad \big| a_n | < \varepsilon \quad\Leftrightarrow\quad \big| a_n - 0| < \varepsilon, $$ which shows that $\lim_{n\to\infty} a_n = 0$. (In fact the argument shows that the two statements are equivalent.)

The same argument generalizes to the other situations you list.

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