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

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

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

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

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

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

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

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

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

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

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

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

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

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

wio
 one year ago
Best ResponseYou've already chosen the best response.1Well 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
 one year ago
Best 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

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

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

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

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