find the limit of 3x^2+6x-9/2x^2-2 as x approaches 1

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find the limit of 3x^2+6x-9/2x^2-2 as x approaches 1

Mathematics
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You might find this site useful: http://archives.math.utk.edu/visual.calculus/1/limits.7/ It has good walk-through tutorials on this topic.
you basically just need to brush up on factoring
(a+b)^2=a^2+2ab+b^2, if you factor a 3 out of that numerator, you will have this situation

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Other answers:

yeah how do i factor 9 to add up to 6
or do i not have to
factor out a 3 first, it will make it much easier
ive got (x+3)(x-3)
thats wrong
thats a difference of 2 squares
so what do i do
if you expanded that you'd get x^2-9
this part 3x^2+6x-9
factor out a 3
3,6,9 all have a common factor of 3 here
3(x^2 +6x-9)
you have to factor the 3 out of the whole numerator so that when you multiply it you get the original expression back
what you just did will yield 3x^2 + 18x -27
3(x^2-2x-3)
+2x
good
now factor the denominator
very much the same thing, guess which number you're pulling out?
2(x^2-1)
yes
now factor (x^2-1)
(difference of 2 squares), i mentioned it a little bit ago
hmm im not sure here
hehe you just did it mistakenly earlier
remember you typed (x-3)(x+3)
oh that
(x+1)(x-1)
yes, now it's all factored. put it all together and cancel like terms
ok but could i have plug in 1 and get an answer
actually, you can further factor the numerator
right now you have the numerator at (x^2+2x-3)
you could not have p lugged 1 into the original equation before factoring it
look at the original denominator
2x^2-2
if you plugged a 1 in here, you'd get a divide by 0 error
i see you will just get 0/0
so try factoring x^2+2x-3
(x-1)(x+3)
good
now everything is factored
so when i cancel i get x+3/x+1
don't forget about the 3 and 2 you factored out
3(x+3)/(2(x+1)
at this point you can plug in 1 and you will not get a divide by 0 error
because you cancelled out the (x-1)
ok i get 12/4
when i reduce i get 3/1
yes. that's it
thanks for the help
3
np
i have another limit problem but its approaching infinity
ok, let's have it
x^2 + 2x-3/2x^2-2 as x approaches infinity, what is the limit
now look at this numerator and denominator
you already factored them in the last problem
give me your factored equation for this problem
(x-1)(x+3)/2(x^2-1)
the denominator further factors out to (x-1)(x+1)
yes, now cancel
x+1/x+1
or 2x+1
what happened to x+3?
my final looks like this (x+3)/(2x+1)
yes
2x+2
no
2(x+1) then you foil
that's as far as you can factor it: (x+3)/[2(x+1)]
you cant do 2x+2?
what's your thinking on that? just trying to see where you're going wrong
oh my mistake, i see what you were doing
thats not a big deal to expand the denominator to 2x+2
so youd have (x+3)/(2x+2)
i thought you meant the whole equation factored out to that
if you have 2(x+1)...why not just foil (2)(x)=2x...2(1)=2...2x+2...
you can, but it's generally not necessary after you've already factored
plus if you do it too early you'll have to factor it again later
oh ok
so since i can't plug in infinity...what should i do
now you have to think what would happen to f(x) is you kept plugging in very large numbers
it would remain large
nope
try plugging in 100 just to test it
well it might cut down...but at the end it would still be large
100+3/2(100+1)=103/202= .50
ok now 1000
the same .50
ok
so 1/2?
1 sec
that's right
ok thanks for the help
sure
just understand why that's the answer
when you start using very large numbers, +3 in the numerator and the +2 in the denominator are pretty much meaningless
essentially its like graphing x/(2x)
ok
thanks for the explanaition

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