1. ## limit proof

Hi all,

Just wondering if I could get help with the following problem.

The question states that I have to prove that:

if lim(x->c) f(x) = L, then lim(x->c)|f(x)| = |L|

I've already proven the above. But then the question goes onto saying.

Show that the converse if false. Give an example where:

lim(x->c)|f(x)| = |L| and lim(x->c)f(x) = M where M is not equal to L.

Also, it asks to give an example where

lim(x->c)|f(x)| exists but lim(x->c)f(x) does not exist.

I'm a bit stumped with this, I can't think of one simple example. Any help would be appreciated.

Cheers.

2. Originally Posted by mosostow
Hi all,

Just wondering if I could get help with the following problem.

The question states that I have to prove that:

if lim(x->c) f(x) = L, then lim(x->c)|f(x)| = |L|

I've already proven the above. But then the question goes onto saying.

Show that the converse if false. Give an example where:

lim(x->c)|f(x)| = |L| and lim(x->c)f(x) = M where M is not equal to L.

Also, it asks to give an example where

lim(x->c)|f(x)| exists but lim(x->c)f(x) does not exist.

I'm a bit stumped with this, I can't think of one simple example. Any help would be appreciated.

.

3. Here is another one I like.
Suppose that f(x)=1 if x is rational and f(x)=-1 if x is irrational. Now |f(x)| has a limit everywhere and it is 1; but f(x) has a limit nowhere.

4. Originally Posted by Plato
Here is another one I like.
Suppose that f(x)=1 if x is rational and f(x)=-1 if x is irrational. Now |f(x)| has a limit everywhere and it is 1; but f(x) has a limit nowhere.
I just love the Dirichlet and his function. I like him so much that I managed to somehow mention his name in my English composition essay.

5. Originally Posted by Plato
Here is another one I like.
Suppose that f(x)=1 if x is rational and f(x)=-1 if x is irrational. Now |f(x)| has a limit everywhere and it is 1; but f(x) has a limit nowhere.
Thanks, this answers the second part of the question.

But I'm still a bit stumped with the first part, that is, to give one example where

if lim(x->c)|f(x)| = |L| then lim(x->c)f(x) = M where M not equal to L.

To me it seems like a logical absurdity but apparently an example does exist.

6. What about f(x) = -1 then lim (x-->c)f(x)=-1

But, lim (x-->c)|f(x)| = 1

And 1 != -1