# Math Help - How to show there's no limit at a jump

1. ## How to show there's no limit at a jump

For the function f:R -->R which is

f(x): -1 when x<0
1 when x>0
0 when x=0

prove that there is no limit at x=0. I'm told I can use a proof by contradiction.

What do you guys think? I've seen I can do something with derivatives or something, but I haven't been taught that yet. I just learned delta-epsilon proofs. (I looked at the sticky, but it wasn't helpful to me).

Thanks!

Regards,
Esther

2. Originally Posted by syok
For the function f:R -->R which is
f(x): -1 when x<0
1 when x>0
0 when x=0
prove that there is no limit at x=0.
It is sufficient to note that $\lim _{x \to 0^ - } f(x) = - 1\;\& \,\lim _{x \to 0^ + } f(x) = 1$

If a limit exists at $x=a$ it in necessary that $\lim _{x \to a^ - } f(x) = \lim _{x \to 0^ + } f(x)$