My question is in regards to number 3 on the below picture. The book wants me to find the points of continuity and the points of discontinuity of the function and to identify each type of discontinuity. My question is how am I supposed to do #3 with limits?
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OpenStudy (anonymous):
Wait the pic was messed up
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
OpenStudy (anonymous):
This one is better
OpenStudy (anonymous):
What grade are you in?
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OpenStudy (anonymous):
11th. AP Calc BC
OpenStudy (anonymous):
Ok
OpenStudy (anonymous):
Does the pic work?
OpenStudy (anonymous):
yeah
OpenStudy (anonymous):
Ok. Now how to do it...
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OpenStudy (anonymous):
@jim_thompson5910
OpenStudy (anonymous):
Wait one sec :T
OpenStudy (anonymous):
Ok, thanks!!!
OpenStudy (anonymous):
Np lol
OpenStudy (anonymous):
I gtg for like an hour.
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OpenStudy (anonymous):
okm
OpenStudy (anonymous):
Ok, I'm back
OpenStudy (anonymous):
What do you already know about limits?
OpenStudy (anonymous):
?
OpenStudy (anonymous):
A lot?
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OpenStudy (anonymous):
I don't know how to prove that this function is continuous. Thats the issue.
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
I can't open that file
OpenStudy (anonymous):
Ok well it says
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OpenStudy (dangerousjesse):
Most people don't download files from students :/
OpenStudy (anonymous):
|dw:1409877512647:dw|
OpenStudy (anonymous):
y=(1)/(x^2+1)
jimthompson5910 (jim_thompson5910):
can the denominator ever be equal to zero?
jimthompson5910 (jim_thompson5910):
ie, are there any solutions to x^2 + 1 = 0 ?
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OpenStudy (anonymous):
Well excluding imaginary numbers, no
jimthompson5910 (jim_thompson5910):
so there are no real numbered values that you can plug into x to make x^2 + 1 = 0 true
that means there is no possible way to divide by zero
OpenStudy (anonymous):
Yes
jimthompson5910 (jim_thompson5910):
visually this makes a graph with no discontinuities
OpenStudy (anonymous):
I know this, but I need to use limits
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jimthompson5910 (jim_thompson5910):
there are no excluded values from the domain
OpenStudy (anonymous):
So there isn't a way to really prove this?
jimthompson5910 (jim_thompson5910):
that's the only way I can think of
OpenStudy (anonymous):
Ok, thanks for the help!
jimthompson5910 (jim_thompson5910):
sure thing
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OpenStudy (anonymous):
Srry I couldn't help you WaffleMan You're really smart Jim