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Say I have a real valued function $f(x)$, is it true that if $\lim_{x \rightarrow c} |f(x)| = 0$ then $\lim_{x \rightarrow c} f(x) = 0$?, where $c$ can be a real number or $\pm \infty$.

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3 Answers 3

up vote 6 down vote accepted

Yes.

Proof. Suppose $\lim\limits_{x \to c} |f(x)| = 0$. Then $\lim\limits_{x \to c} -|f(x)| = 0$ also. For any $x$, we have $$ -|f(x)| \le f(x) \le |f(x)| $$ implying $\lim\limits_{x \to c} f(x) = 0$ by the squeeze theorem.


Note that $\lim_{x \to c} |f(x)| = L$ does not imply the existence of $\lim_{x \to c} f(x)$ when $L \ne 0$.

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Well, $\lim_{x \rightarrow c} |f(x)| = L$ implies that if $\lim_{x \rightarrow c} f(x)$ exists, it is either $L$ or $-L$, no? –  Istvan Chung May 17 '14 at 20:56
    
@IstvanChung Yes, that is true. "anything about" is poor wording; I'll change it. –  C-S May 17 '14 at 23:42
    
I think this is not true take $f(n)=(-1)^n$. While $\lim_{n\to \infty}|f(n)|=1$ but $f(n)$ does not have a limit at infinity. –  palio Mar 23 at 11:24
    
@palio If you are speaking about Istvan's comment, reread it. One of the hypotheses is that the limit exist. –  C-S Mar 24 at 7:01
    
@C-S yes i reread it. Thanks !! –  palio Mar 25 at 3:35

Yes

Try

$|f(x)|= ||f(x)|-0| < \epsilon $

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Yes. Note that $\lim_{x\to c}f(x)=0$ translates to: For all $\epsilon>0$ there exists a $\delta>0$ such that for all $x$ with $0<|x-c|<\delta$ (or accordingly for the infinite case) we have $|f(x)-0|<\epsilon$. But clearly $$ |f(x)-0|<\epsilon\iff |f(x)|<\epsilon\iff \bigl||f(x)|-0\bigr|<\epsilon.$$ Hence the condition for $f$ and $|f|$ is in fact the same.

Note that for any nonzero limit $a$ we'd only have the other direction $$ |f(x)-a|<\epsilon\implies\bigl||f(x)|-|a|\bigr|<\epsilon.$$

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