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

- anonymous

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

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- schrodinger

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- 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

you basically just need to brush up on factoring

- 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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## More answers

- anonymous

yeah how do i factor 9 to add up to 6

- anonymous

or do i not have to

- anonymous

factor out a 3 first, it will make it much easier

- anonymous

ive got (x+3)(x-3)

- anonymous

thats wrong

- anonymous

thats a difference of 2 squares

- anonymous

so what do i do

- anonymous

if you expanded that you'd get x^2-9

- anonymous

this part 3x^2+6x-9

- anonymous

factor out a 3

- anonymous

3,6,9 all have a common factor of 3 here

- anonymous

3(x^2 +6x-9)

- 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

what you just did will yield 3x^2 + 18x -27

- anonymous

3(x^2-2x-3)

- anonymous

+2x

- anonymous

good

- anonymous

now factor the denominator

- anonymous

very much the same thing, guess which number you're pulling out?

- anonymous

2(x^2-1)

- anonymous

yes

- anonymous

now factor (x^2-1)

- anonymous

(difference of 2 squares), i mentioned it a little bit ago

- anonymous

hmm im not sure here

- anonymous

hehe you just did it mistakenly earlier

- anonymous

remember you typed (x-3)(x+3)

- anonymous

oh that

- anonymous

(x+1)(x-1)

- anonymous

yes, now it's all factored. put it all together and cancel like terms

- anonymous

ok but could i have plug in 1 and get an answer

- anonymous

actually, you can further factor the numerator

- anonymous

right now you have the numerator at (x^2+2x-3)

- anonymous

you could not have p lugged 1 into the original equation before factoring it

- anonymous

look at the original denominator

- anonymous

2x^2-2

- anonymous

if you plugged a 1 in here, you'd get a divide by 0 error

- anonymous

i see you will just get 0/0

- anonymous

so try factoring x^2+2x-3

- anonymous

(x-1)(x+3)

- anonymous

good

- anonymous

now everything is factored

- anonymous

so when i cancel i get x+3/x+1

- anonymous

don't forget about the 3 and 2 you factored out

- anonymous

3(x+3)/(2(x+1)

- anonymous

at this point you can plug in 1 and you will not get a divide by 0 error

- anonymous

because you cancelled out the (x-1)

- anonymous

ok i get 12/4

- anonymous

when i reduce i get 3/1

- anonymous

yes. that's it

- anonymous

thanks for the help

- anonymous

3

- anonymous

np

- anonymous

i have another limit problem but its approaching infinity

- anonymous

ok, let's have it

- anonymous

x^2 + 2x-3/2x^2-2 as x approaches infinity, what is the limit

- anonymous

now look at this numerator and denominator

- anonymous

you already factored them in the last problem

- anonymous

give me your factored equation for this problem

- anonymous

(x-1)(x+3)/2(x^2-1)

- anonymous

the denominator further factors out to (x-1)(x+1)

- anonymous

yes, now cancel

- anonymous

x+1/x+1

- anonymous

or 2x+1

- anonymous

what happened to x+3?

- anonymous

my final looks like this (x+3)/(2x+1)

- anonymous

yes

- anonymous

2x+2

- anonymous

no

- anonymous

2(x+1) then you foil

- anonymous

that's as far as you can factor it: (x+3)/[2(x+1)]

- anonymous

you cant do 2x+2?

- anonymous

what's your thinking on that? just trying to see where you're going wrong

- anonymous

oh my mistake, i see what you were doing

- anonymous

thats not a big deal to expand the denominator to 2x+2

- anonymous

so youd have (x+3)/(2x+2)

- anonymous

i thought you meant the whole equation factored out to that

- anonymous

if you have 2(x+1)...why not just foil (2)(x)=2x...2(1)=2...2x+2...

- anonymous

you can, but it's generally not necessary after you've already factored

- anonymous

plus if you do it too early you'll have to factor it again later

- anonymous

oh ok

- anonymous

so since i can't plug in infinity...what should i do

- anonymous

now you have to think what would happen to f(x) is you kept plugging in very large numbers

- anonymous

it would remain large

- anonymous

nope

- anonymous

try plugging in 100 just to test it

- anonymous

well it might cut down...but at the end it would still be large

- anonymous

100+3/2(100+1)=103/202= .50

- anonymous

ok now 1000

- anonymous

the same .50

- anonymous

ok

- anonymous

so 1/2?

- anonymous

1 sec

- anonymous

that's right

- anonymous

ok thanks for the help

- anonymous

sure

- anonymous

just understand why that's the answer

- anonymous

when you start using very large numbers, +3 in the numerator and the +2 in the denominator are pretty much meaningless

- anonymous

essentially its like graphing x/(2x)

- anonymous

ok

- anonymous

thanks for the explanaition

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