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Use numerical and graphical evidence to conjecture value(s) for the limits, if they exist. If not, describe what is happening at x = a graphically.

Mathematics
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1) lim x-> 1 (x^2 - 1)/(x + 2) 2) lim x-> 1 (x^2 + 1)/(x - 1)
@UnkleRhaukus umm can u help?
1) \[\lim\limits_ {x\to1}\frac{ x^2 - 1}{x + 2}\] what happens when we put x=1 into the fraction

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Other answers:

hmm its becomes 0/3?
yeah, do you know what that simplifies to ?
umm 0?
correct,
okay so how would i describe it?
can you fill in the table |dw:1359329822568:dw|
yes i can
okay let me try this
so for the -2 one i'm dividing -5 to 0 so will it be -5/0 since i cant divide that?
ok so i got |dw:1359330562610:dw|
so did i get it?
|dw:1359330502446:dw|
are u there?
ok
how about now? |dw:1359331215777:dw|
umm are u there?
(sorry my internet cut out for a moment there) um you got most of them right the first time but check the case when x=-1 agian |dw:1359331049113:dw|
okay
so then i draw the graph? right ?
yeah not plot those points |dw:1359331297137:dw|
not*now
ok
|dw:1359331448238:dw| maybe i should have chosen some better x points
what are the question marks and the lines?
ive put question marks for for the x=-2, y=-5/0 bit
ohh okay
|dw:1359331760380:dw|
so its a curve?
if i chose better x values we would have got something like this
okay
so that will be the end of the problem?
|dw:1359332271558:dw|
|dw:1359332425565:dw|
|dw:1359332578040:dw|
ok so this is the final graph?
this shape is called a hyperbola at x=-2 y is undefined, this is because if we look at the graph is should be +∞ and also -∞  but it can't be more than one point we call this a vertical Asymptote
okay
|dw:1359332784659:dw|
ohh ok i get it
to describe what is happening at x=a if \(a < -2\) the value will be on the bottom curve if \(a = -2\) the value is undefined if \(a >2\) the value will be on the top curve
__________ for 2) \[ \lim\limits_{x\to 1} \frac{x^2 + 1}{x - 1} \] the value is undefined as we are trying to taking the limit at the asymptote
okay
i hope i haven't confused you too much
noo u havent so i'll do the same this with lim x-> 1 (x^2 + 1)/(x -1)
another way of saying the value is undefined is to say the limit does not exist
okay
so i'll do the same for lim x-> 1 (x^2 + 1)/(x -1)?
yeah, i would use x =-5,-4,-3,-2,-1,0,2,3,4,5 this time
ok thanks!! :)
ok im off, good luck
hey so i did it is this correct? |dw:1359335654116:dw| |dw:1359335759751:dw|
@satellite73 can u help with this?

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