# If the $\lim(|f(x)|) = 0$ then $\lim(f(x)) = 0$. Does this work both ways?

We were given a list of limit laws in our Calculus Study Guide and I can't understand why this was given as a one-way thing.

GIVEN:

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

This is the limit property that we have been given. Now, why can't we say:

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

I can't think of any graphs I know of that would make the above false. Can anybody explain why this only works one way?

Excuse my mathematically-criminal language; I am busy preparing for a test at the moment.

• The part you are referring to is automatically true. You can see this directly from the $\epsilon$, $\delta$ definition. Another method is to use the continuity of the absolute value function at 0. – ayberk Mar 14 '17 at 11:41
• Worked out in $\epsilon-\delta$ language the two statements are exactly the same. This because $||f(x)||=|f(x)|$. – drhab Mar 14 '17 at 11:42
• Thanks! Our lecturer said she would think of an example to prove the converse wrong after somebody asked but she never followed through with that. – Delonjnaidu Mar 14 '17 at 11:48

$\lim_{x \to a}f(x)=0$ $\Leftrightarrow$ $\forall \varepsilon >0, \exists \delta>0 ; 0<|x-a|<\delta \Rightarrow |f(x)-0|=|f(x)|<\varepsilon$ $\Leftrightarrow$ $\forall \varepsilon >0, \exists \delta>0 ; 0<|x-a|<\delta \Rightarrow ||f(x)|-0|=|f(x)|<\varepsilon$
As $|f(x)-0|=||f(x)|-0|$, the computation of the limit will work both ways.