#1

k basically i need to know how to do the following:

lim when x approaches 1 for: (x^3-4x^2+3)/(x^2-1)

here i dont really know how to factor the numerator.

lim when x approaches -infinite. (3-10x)/(square root of(25x^2-4x+1))

here i get 3/2 as an answer but its supposed to be -3/2. its something to do with the -infinite but i dont understand.

last one:

how do i factor: (x^2)/(square root of(x^2+4)-2)

thanks for any help. really appreciated

edit: NEW QUESTION YAY!!

and a new question im going to add:

we have this equation:

(e^(-x)e^(-2x)e^(4x-x^2)

apparently this is equal to e^(x-x^3)

but i thought you were supposed to just add the exponants so it would be e^(x-x^2)

lim when x approaches 1 for: (x^3-4x^2+3)/(x^2-1)

here i dont really know how to factor the numerator.

lim when x approaches -infinite. (3-10x)/(square root of(25x^2-4x+1))

here i get 3/2 as an answer but its supposed to be -3/2. its something to do with the -infinite but i dont understand.

last one:

how do i factor: (x^2)/(square root of(x^2+4)-2)

thanks for any help. really appreciated

edit: NEW QUESTION YAY!!

and a new question im going to add:

we have this equation:

(e^(-x)e^(-2x)e^(4x-x^2)

apparently this is equal to e^(x-x^3)

but i thought you were supposed to just add the exponants so it would be e^(x-x^2)

*Last edited by Teh_Guitarer at Oct 28, 2007,*

#2

beats me.

#3

how do i factor: (x^2)/(square root of(x^2+4)-2)

don't u multiply the numerator and denominator by the w/e it's called of the denominator?

don't u multiply the numerator and denominator by the w/e it's called of the denominator?

#4

how do i factor: (x^2)/(square root of(x^2+4)-2)

don't u multiply the numerator and denominator by the w/e it's called of the denominator?

ahhhhhhhhhh right. you're good

thanks alot man

#5

firrst one is -5/2

when u put in one u get 0 over 0 an indeterminate form.

U can use L'Hopital's Rule and take the derivative of the top and bottom and then factor the x back in you should get (3x^2-8x)/(2x)

(3(1)^2-8(1))/(2(1))

(3-8)/2

-5/2

when u put in one u get 0 over 0 an indeterminate form.

U can use L'Hopital's Rule and take the derivative of the top and bottom and then factor the x back in you should get (3x^2-8x)/(2x)

(3(1)^2-8(1))/(2(1))

(3-8)/2

-5/2

#6

for the first one: the numerator factors into (X^2-1)(X-3)

then you can cancel the (X-3) from the numerator with the one from the denominator

edit: the guy above me is right...i didn't think to check to see if it was indeterminate form

then you can cancel the (X-3) from the numerator with the one from the denominator

edit: the guy above me is right...i didn't think to check to see if it was indeterminate form

#7

^haha hes better than me.

my calculus knowledge is from two years ago junior year.

my calculus knowledge is from two years ago junior year.

#8

Calc BC this year and im a senior in high school

#9

i took AP calc AB junior year. and slacked off in statistics senior year.

#10

yeah stat is ****

#11

Taking AP Calc AB right now, and I have no idea...we're only doing the Chain Rule right now

#12

for the first one: the numerator factors into (X^2-1)(X-3)

then you can cancel the (X-3) from the numerator with the one from the denominator

edit: the guy above me is right...i didn't think to check to see if it was indeterminate form

how did you factor the numerator? thanks

#13

i just factored (X^3-4X+3)how did you factor the numerator? thanks

i figured that (X^2-1) would be one of the factors since it was in the denominator so i started with that

it's in indeterminate form though so you need to do what the other guy said

#14

Taking AP Calc AB right now, and I have no idea...we're only doing the Chain Rule right now

we're doing that in physics... its so hard after taking a break with statistics for a year... i forgot everything.

#15

alright thanks. anyone know number 2? and a new question im going to add:

we have this equation:

(e^(-x)e^(-2x)e^(4x-x^2)

apparently this is equal to e^(x-x^3)

but i thought you were supposed to just add the exponants so it would be e^(x-x^2)

am i wrong?

we have this equation:

(e^(-x)e^(-2x)e^(4x-x^2)

apparently this is equal to e^(x-x^3)

but i thought you were supposed to just add the exponants so it would be e^(x-x^2)

am i wrong?