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3psilon Group TitleBest ResponseYou've already chosen the best response.0
\[f(x) = (x ^{2}1) x \neq 1 \]\[f(x) = (4) x = 1 \]
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
f(x) = 4 at x = 1
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
It's a polynomial, meaning it is continuous for all real numbers meaning you can just plug it in to find the limit.
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
dw:1361493037294:dw
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
So it still exists?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Wait, so we have \[ f(x) = \begin{cases} x^21 & x\neq 1\\ \text{undefined} & x=1 \end{cases} \]Right?
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Yes I just couldn't get it to look like that. Undefined should be 4 though .
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
So it's just \[ \lim_{x\to1}f(x) = (1)^21 \]
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
We can get really really close to 1 and we'll approach 0
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Do you understand the epsilon delta definition of limit?
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
On the graph it's not continuous but does the limit still exist is what I'm asking
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Yes, it has just been a while since i've done limits
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
When you want to know if a limit exists, just look at the right hand and left hand limits.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
If they meet, then the limit exists.
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
But @wio They don't meet at that point cause at X=1 the point is shifted up
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
What are you getting for the left and right limits?
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
I don't know what to do because the point isn't there. It's shifted up
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
@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
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Did you plug in numbers getting close to 1 on both sides? What are they approaching?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
No, it is approaching 0.
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
So the limit does exist as it approaches 1?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
\((0.9999)^21 \approx 0\)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
actually\[f(x) = \begin{cases} x^21 & x\neq 1\\ 4 & x=1 \end{cases}\]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.
 one year ago
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