3psilon
  • 3psilon
Does this limit exists as x-> 1 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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3psilon
  • 3psilon
\[f(x) = (x ^{2}-1) x \neq 1 \]\[f(x) = (4) x = 1 \]
3psilon
  • 3psilon
f(x) = 4 at x = 1
anonymous
  • 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
  • 3psilon
|dw:1361493037294:dw|
3psilon
  • 3psilon
So it still exists?
anonymous
  • anonymous
Wait, so we have \[ f(x) = \begin{cases} x^2-1 & x\neq 1\\ \text{undefined} & x=1 \end{cases} \]Right?
3psilon
  • 3psilon
Yes I just couldn't get it to look like that. Undefined should be 4 though .
anonymous
  • anonymous
So it's just \[ \lim_{x\to1}f(x) = (1)^2-1 \]
3psilon
  • 3psilon
yes
anonymous
  • anonymous
We can get really really close to 1 and we'll approach 0
anonymous
  • anonymous
Do you understand the epsilon delta definition of limit?
3psilon
  • 3psilon
On the graph it's not continuous but does the limit still exist is what I'm asking
3psilon
  • 3psilon
Yes, it has just been a while since i've done limits
anonymous
  • anonymous
When you want to know if a limit exists, just look at the right hand and left hand limits.
anonymous
  • anonymous
If they meet, then the limit exists.
3psilon
  • 3psilon
But @wio They don't meet at that point cause at X=1 the point is shifted up
anonymous
  • anonymous
What are you getting for the left and right limits?
3psilon
  • 3psilon
I don't know what to do because the point isn't there. It's shifted up
anonymous
  • 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
  • 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
  • anonymous
Did you plug in numbers getting close to 1 on both sides? What are they approaching?
3psilon
  • 3psilon
4
anonymous
  • anonymous
No, it is approaching 0.
3psilon
  • 3psilon
So the limit does exist as it approaches 1?
anonymous
  • anonymous
\((0.9999)^2-1 \approx 0\)
anonymous
  • anonymous
Yeah.
TuringTest
  • 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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