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Does this limit exists as x-> 1 ?

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\[f(x) = (x ^{2}-1) x \neq 1 \]\[f(x) = (4) x = 1 \]
f(x) = 4 at x = 1
It's a polynomial, meaning it is continuous for all real numbers meaning you can just plug it in to find the limit.

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Other answers:

So it still exists?
Wait, so we have \[ f(x) = \begin{cases} x^2-1 & x\neq 1\\ \text{undefined} & x=1 \end{cases} \]Right?
Yes I just couldn't get it to look like that. Undefined should be 4 though .
So it's just \[ \lim_{x\to1}f(x) = (1)^2-1 \]
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.
But @wio They don't meet at that point cause at X=1 the point is shifted up
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
Well if you wanted to find the right hand limit, you're going to plug in things like .999 and when you wanted to find the left hand limit, you're going to plug in things like 1.0001
@wio this is not helping at all. Can you please just answer my question if it exists or not. Not trying to be rude but I'd appreciate it
Did you plug in numbers getting close to 1 on both sides? What are they approaching?
No, it is approaching 0.
So the limit does exist as it approaches 1?
\((0.9999)^2-1 \approx 0\)
actually\[f(x) = \begin{cases} x^2-1 & x\neq 1\\ 4 & x=1 \end{cases}\]but that is not important. if all we are asking is if the limit exists, then @wio is heading you in the right direction. only if they are asking if it is continuous does the other part matter.

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