onegirl
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.
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onegirl
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1) lim x-> 1 (x^2 - 1)/(x + 2)
2) lim x-> 1 (x^2 + 1)/(x - 1)
onegirl
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@UnkleRhaukus umm can u help?
UnkleRhaukus
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1) \[\lim\limits_ {x\to1}\frac{ x^2 - 1}{x + 2}\]
what happens when we put x=1 into the fraction
onegirl
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hmm its becomes 0/3?
UnkleRhaukus
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yeah, do you know what that simplifies to ?
onegirl
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umm 0?
UnkleRhaukus
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correct,
onegirl
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okay so how would i describe it?
UnkleRhaukus
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can you fill in the table |dw:1359329822568:dw|
onegirl
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yes i can
onegirl
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okay let me try this
onegirl
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so for the -2 one i'm dividing -5 to 0 so will it be -5/0 since i cant divide that?
onegirl
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ok so i got |dw:1359330562610:dw|
onegirl
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so did i get it?
UnkleRhaukus
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|dw:1359330502446:dw|
onegirl
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are u there?
onegirl
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ok
onegirl
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how about now? |dw:1359331215777:dw|
onegirl
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umm are u there?
UnkleRhaukus
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(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|
onegirl
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okay
onegirl
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so then i draw the graph? right ?
UnkleRhaukus
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yeah not plot those points |dw:1359331297137:dw|
UnkleRhaukus
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not*now
onegirl
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ok
UnkleRhaukus
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|dw:1359331448238:dw|
maybe i should have chosen some better x points
onegirl
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what are the question marks and the lines?
UnkleRhaukus
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ive put question marks for for the x=-2, y=-5/0 bit
onegirl
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ohh okay
UnkleRhaukus
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|dw:1359331760380:dw|
onegirl
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so its a curve?
UnkleRhaukus
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if i chose better x values
we would have got something like this
onegirl
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okay
onegirl
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so that will be the end of the problem?
UnkleRhaukus
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|dw:1359332271558:dw|
UnkleRhaukus
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|dw:1359332425565:dw|
UnkleRhaukus
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|dw:1359332578040:dw|
onegirl
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ok so this is the final graph?
UnkleRhaukus
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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
onegirl
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okay
UnkleRhaukus
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|dw:1359332784659:dw|
onegirl
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ohh ok i get it
UnkleRhaukus
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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
UnkleRhaukus
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__________
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
onegirl
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okay
UnkleRhaukus
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i hope i haven't confused you too much
onegirl
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noo u havent so i'll do the same this with lim x-> 1 (x^2 + 1)/(x -1)
UnkleRhaukus
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another way of saying the value is undefined is to say the limit does not exist
onegirl
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okay
onegirl
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so i'll do the same for lim x-> 1 (x^2 + 1)/(x -1)?
UnkleRhaukus
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yeah, i would use
x =-5,-4,-3,-2,-1,0,2,3,4,5
this time
onegirl
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ok thanks!! :)
UnkleRhaukus
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ok im off, good luck
onegirl
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hey so i did it is this correct? |dw:1359335654116:dw| |dw:1359335759751:dw|
onegirl
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@satellite73 can u help with this?
onegirl
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@zepdrix ??
onegirl
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@jim_thompson5910 ??