Does this limit exists as x-> 1 ?

- 3psilon

Does this limit exists as x-> 1 ?

- jamiebookeater

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- 3psilon

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

- 3psilon

f(x) = 4 at x = 1

- anonymous

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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## More answers

- 3psilon

|dw:1361493037294:dw|

- 3psilon

So it still exists?

- anonymous

Wait, so we have \[
f(x) = \begin{cases}
x^2-1 & x\neq 1\\
\text{undefined} & x=1
\end{cases}
\]Right?

- 3psilon

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

- anonymous

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

- 3psilon

yes

- anonymous

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

- anonymous

Do you understand the epsilon delta definition of limit?

- 3psilon

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

- 3psilon

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

- anonymous

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

- anonymous

If they meet, then the limit exists.

- 3psilon

But @wio They don't meet at that point cause at X=1 the point is shifted up

- anonymous

What are you getting for the left and right limits?

- 3psilon

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

- anonymous

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

- 3psilon

@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

- anonymous

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

- 3psilon

4

- anonymous

No, it is approaching 0.

- 3psilon

So the limit does exist as it approaches 1?

- anonymous

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

- anonymous

Yeah.

- TuringTest

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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