Question

Limit Question \[
\lim_{x\to 0}f(|x|) = \lim_{x\to0^+}f(x)
\]I am wondering if this is true or not.

Answer 1

I mean \(f(x) = x\), we have: \[
\lim_{x\to 0}|x| =0= \lim_{x\to 0^+}x
\]But I'm wondering about more general cases.

Answer 2

\(\lim_{x\to 0}|x| =0= \lim_{x\to 0^-}x\)
also ...

and for
\(\lim_{x\to 0}f(|x|) \)
to exist

it must be
\( = \lim_{x\to0^+}f(x)= \lim_{x\to0^-}f(x)\)

both

Answer 3

Really @hartnn ?

Answer 4

I think this is true:\[
\lim_{x\to 0}\frac{1}{|x|} = \infty
\]

Answer 5

I'm using the function \(f(x) = 1/x\) this time.

Answer 6

But doesn't: \[
\lim_{x\to 0^-}\frac{1}{x} = -\infty
\]?

Answer 7

Or did I mix them up?

Answer 8

yes wio,
and yes thats true
even for f(x) =1/x
or any f(x)

if lim x->0 f(|x|) exist,
then its a must that
\(= \lim_{x\to0^+}f(x)= \lim_{x\to0^-}f(x) =\lim_{x\to0}f(x) \)

Answer 9

if you are taking x->0-

then |x| = -x

Answer 10

\(\lim_{x\to 0^-}\frac{1}{-x} = \infty\)

Answer 11

Why do both limits have to exist?

Answer 12

woah!
why are so many offline memebers atching this post :O :P

Answer 13

*watching
and they keep on increasing !

Answer 14

Since \(|x|>0\) mean that the negative side doesn't matter?

Answer 15

@hartnn Why does the negative side matter if \(|x|>0\)?

Answer 16

my guess : some school teacher introduced OpenStudy to their class,
and all of them came to check it out without logging in :P

wio,
f(x) = 1/x
f(|x|) = 1/|x|

for x->0+ , x>0
1/|x| = 1/x

for x->0-,
x<0
so 1/|x| = -1/x

Answer 17

\[
\lim_{x\to 0}f(|x|)=L
\]Okay so we have: \[
|x|<\delta \implies |f(|x|)-L)|<\epsilon
\]

Answer 18

@hartnn Yeah, but in this case \(f(|x|)\) exists but \(f(x)\) doesn't. So that already is against what you said.

Answer 19

you mean at x=0 ?

Answer 20

Yes

Answer 21

You said that \(f(|x|)\) doesn't exist unless \(f(x)\) does, but that doesn't seem to be true for \(f(x)=1/x\).

Answer 22

i thought,
f(x) exist at x=0 and is = \(\infty \)

Answer 23

For \[
\lim_{x\to0+} f(x)=L
\]We have: \[
x<\delta \implies |f(x)-L|<\epsilon
\]

Answer 24

And \(0<x\) as well

Answer 25

Doesn't it seem like they're equivalent?

Answer 26

If you let \(y=|x|\), then they are equivalent, I think

Answer 27

@hartnn Do you want to understand delta epsilon a bit better?

Answer 28

yup, f(x) does not exist at x=0

also i meant
\(= \lim_{x\to0^+}|f(x)|= \lim_{x\to0^-}|f(x)|\)

i just saw i missed || in my reply

i never meant " f(|x|) doesn't exist unless f(x) does, but that doesn't seem to be true for f(x)=1/x."

Answer 29

\(\color{blue}{\text{Originally Posted by}}\) @hartnn
if lim x->0 f(|x|) exist,
then its a must that
\(= \lim_{x\to0^+}f(x)= \lim_{x\to0^-}f(x) =\lim_{x\to0}f(x) \)
\(\color{blue}{\text{End of Quote}}\)
This is tha part that confused me...

Answer 30

yes, i missed || there too :(
\(= \lim_{x\to0^+}|f(x)|= \lim_{x\to0^-}|f(x)| =\lim_{x\to0}|f(x)|\)

Answer 31

Oh, ok. I was thinking about doing a delta epsilon tutorial

Answer 32

go on!
many users will be benefited, including me :)

Answer 33

Okay, I think I'll write it up in a new question, but I probably better plan it out a bit

Answer 34

did we conclude on your question ?

Answer 35

Yeah, I'll close this.

Answer 36

I'm just using this to preview mathjax