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\[f(x) = (x ^{2}-1) x \neq 1 \]\[f(x) = (4) x = 1 \]

f(x) = 4 at x = 1

|dw:1361493037294:dw|

So it still exists?

Yes I just couldn't get it to look like that. Undefined should be 4 though .

So it's just \[
\lim_{x\to1}f(x) = (1)^2-1
\]

yes

We can get really really close to 1 and we'll approach 0

Do you understand the epsilon delta definition of limit?

On the graph it's not continuous but does the limit still exist is what I'm asking

Yes, it has just been a while since i've done limits

When you want to know if a limit exists, just look at the right hand and left hand limits.

If they meet, then the limit exists.

What are you getting for the left and right limits?

I don't know what to do because the point isn't there. It's shifted up

Did you plug in numbers getting close to 1 on both sides? What are they approaching?

4

No, it is approaching 0.

So the limit does exist as it approaches 1?

\((0.9999)^2-1 \approx 0\)

Yeah.