anonymous
  • anonymous
find the limit of 3x^2+6x-9/2x^2-2 as x approaches 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
asnaseer
  • asnaseer
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.
anonymous
  • anonymous
you basically just need to brush up on factoring
anonymous
  • anonymous
(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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anonymous
  • anonymous
yeah how do i factor 9 to add up to 6
anonymous
  • anonymous
or do i not have to
anonymous
  • anonymous
factor out a 3 first, it will make it much easier
anonymous
  • anonymous
ive got (x+3)(x-3)
anonymous
  • anonymous
thats wrong
anonymous
  • anonymous
thats a difference of 2 squares
anonymous
  • anonymous
so what do i do
anonymous
  • anonymous
if you expanded that you'd get x^2-9
anonymous
  • anonymous
this part 3x^2+6x-9
anonymous
  • anonymous
factor out a 3
anonymous
  • anonymous
3,6,9 all have a common factor of 3 here
anonymous
  • anonymous
3(x^2 +6x-9)
anonymous
  • anonymous
you have to factor the 3 out of the whole numerator so that when you multiply it you get the original expression back
anonymous
  • anonymous
what you just did will yield 3x^2 + 18x -27
anonymous
  • anonymous
3(x^2-2x-3)
anonymous
  • anonymous
+2x
anonymous
  • anonymous
good
anonymous
  • anonymous
now factor the denominator
anonymous
  • anonymous
very much the same thing, guess which number you're pulling out?
anonymous
  • anonymous
2(x^2-1)
anonymous
  • anonymous
yes
anonymous
  • anonymous
now factor (x^2-1)
anonymous
  • anonymous
(difference of 2 squares), i mentioned it a little bit ago
anonymous
  • anonymous
hmm im not sure here
anonymous
  • anonymous
hehe you just did it mistakenly earlier
anonymous
  • anonymous
remember you typed (x-3)(x+3)
anonymous
  • anonymous
oh that
anonymous
  • anonymous
(x+1)(x-1)
anonymous
  • anonymous
yes, now it's all factored. put it all together and cancel like terms
anonymous
  • anonymous
ok but could i have plug in 1 and get an answer
anonymous
  • anonymous
actually, you can further factor the numerator
anonymous
  • anonymous
right now you have the numerator at (x^2+2x-3)
anonymous
  • anonymous
you could not have p lugged 1 into the original equation before factoring it
anonymous
  • anonymous
look at the original denominator
anonymous
  • anonymous
2x^2-2
anonymous
  • anonymous
if you plugged a 1 in here, you'd get a divide by 0 error
anonymous
  • anonymous
i see you will just get 0/0
anonymous
  • anonymous
so try factoring x^2+2x-3
anonymous
  • anonymous
(x-1)(x+3)
anonymous
  • anonymous
good
anonymous
  • anonymous
now everything is factored
anonymous
  • anonymous
so when i cancel i get x+3/x+1
anonymous
  • anonymous
don't forget about the 3 and 2 you factored out
anonymous
  • anonymous
3(x+3)/(2(x+1)
anonymous
  • anonymous
at this point you can plug in 1 and you will not get a divide by 0 error
anonymous
  • anonymous
because you cancelled out the (x-1)
anonymous
  • anonymous
ok i get 12/4
anonymous
  • anonymous
when i reduce i get 3/1
anonymous
  • anonymous
yes. that's it
anonymous
  • anonymous
thanks for the help
anonymous
  • anonymous
3
anonymous
  • anonymous
np
anonymous
  • anonymous
i have another limit problem but its approaching infinity
anonymous
  • anonymous
ok, let's have it
anonymous
  • anonymous
x^2 + 2x-3/2x^2-2 as x approaches infinity, what is the limit
anonymous
  • anonymous
now look at this numerator and denominator
anonymous
  • anonymous
you already factored them in the last problem
anonymous
  • anonymous
give me your factored equation for this problem
anonymous
  • anonymous
(x-1)(x+3)/2(x^2-1)
anonymous
  • anonymous
the denominator further factors out to (x-1)(x+1)
anonymous
  • anonymous
yes, now cancel
anonymous
  • anonymous
x+1/x+1
anonymous
  • anonymous
or 2x+1
anonymous
  • anonymous
what happened to x+3?
anonymous
  • anonymous
my final looks like this (x+3)/(2x+1)
anonymous
  • anonymous
yes
anonymous
  • anonymous
2x+2
anonymous
  • anonymous
no
anonymous
  • anonymous
2(x+1) then you foil
anonymous
  • anonymous
that's as far as you can factor it: (x+3)/[2(x+1)]
anonymous
  • anonymous
you cant do 2x+2?
anonymous
  • anonymous
what's your thinking on that? just trying to see where you're going wrong
anonymous
  • anonymous
oh my mistake, i see what you were doing
anonymous
  • anonymous
thats not a big deal to expand the denominator to 2x+2
anonymous
  • anonymous
so youd have (x+3)/(2x+2)
anonymous
  • anonymous
i thought you meant the whole equation factored out to that
anonymous
  • anonymous
if you have 2(x+1)...why not just foil (2)(x)=2x...2(1)=2...2x+2...
anonymous
  • anonymous
you can, but it's generally not necessary after you've already factored
anonymous
  • anonymous
plus if you do it too early you'll have to factor it again later
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
so since i can't plug in infinity...what should i do
anonymous
  • anonymous
now you have to think what would happen to f(x) is you kept plugging in very large numbers
anonymous
  • anonymous
it would remain large
anonymous
  • anonymous
nope
anonymous
  • anonymous
try plugging in 100 just to test it
anonymous
  • anonymous
well it might cut down...but at the end it would still be large
anonymous
  • anonymous
100+3/2(100+1)=103/202= .50
anonymous
  • anonymous
ok now 1000
anonymous
  • anonymous
the same .50
anonymous
  • anonymous
ok
anonymous
  • anonymous
so 1/2?
anonymous
  • anonymous
1 sec
anonymous
  • anonymous
that's right
anonymous
  • anonymous
ok thanks for the help
anonymous
  • anonymous
sure
anonymous
  • anonymous
just understand why that's the answer
anonymous
  • anonymous
when you start using very large numbers, +3 in the numerator and the +2 in the denominator are pretty much meaningless
anonymous
  • anonymous
essentially its like graphing x/(2x)
anonymous
  • anonymous
ok
anonymous
  • anonymous
thanks for the explanaition

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