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3psilon
 3 years ago
Does this limit exists as x> 1 ?
3psilon
 3 years ago
Does this limit exists as x> 1 ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's a polynomial, meaning it is continuous for all real numbers meaning you can just plug it in to find the limit.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait, so we have \[ f(x) = \begin{cases} x^21 & x\neq 1\\ \text{undefined} & x=1 \end{cases} \]Right?

3psilon
 3 years 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 .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So it's just \[ \lim_{x\to1}f(x) = (1)^21 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We can get really really close to 1 and we'll approach 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you understand the epsilon delta definition of limit?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When you want to know if a limit exists, just look at the right hand and left hand limits.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If they meet, then the limit exists.

3psilon
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What are you getting for the left and right limits?

3psilon
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did you plug in numbers getting close to 1 on both sides? What are they approaching?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, it is approaching 0.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\((0.9999)^21 \approx 0\)

TuringTest
 3 years 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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