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CALCULUS HELP!!!

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sstricky
Posts: 1876 - Karma: 111
i have a problem and have no idea how to do it. I NEED FAST!!

lim_________3x^2 - 5x^3 + 4
x -> (infinity) 3x^2 + 5x - 2

whats the limit?!?!
Sep 3 2009 1:10PM
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steevner
Posts: 4502 - Karma: 47
I used to know that stuff. Wow, I forgot a LOT over the summer.
Sep 3 2009 1:21PM
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eheath
Posts: 61110 - Karma: 118,329
lim_________3x^2 - 5x^3 + 4
x -> (infinity) 3x^2 + 5x - 2

so its infinity - infinity + 4 over infinity + infinity - 2
so
4
inifinty-2

And then that means something, i can remember exactly but a number over infinity is basically to 0? hope that helps.
Sep 3 2009 1:22PM
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sstricky
Posts: 1876 - Karma: 111
2389rq3rjowr
Sep 3 2009 1:40PM
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ORski21
Posts: 423 - Karma: 12
okay, i dont think any ones right so far. sorry eheath.

lim x-> infinity for (3x^2 - 5x^3 + 4)/(3x^2 + 5x - 2)

when you plug infinty in for x you get infinity over infinity, and that cant be the answer. so you have to the the derivative of the top and the bottom.

It will be lim as x -> infinty for(6x - 15x^2)/(6x + 5)

this will also be infinity over infinity so you have to take the derivative again.

lim as x-> infinity for (6 - 30x)/(6)

okay, now when you plug infinity into this you get (6 - 30(infinity))/(6), so the answer is negative infinity. I'm pretty sure thats right and hopefully that makes sense.
Sep 3 2009 1:50PM
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eheath
Posts: 61110 - Karma: 118,329
Ah yes, simplify of course. that all sounds much better.
Sep 3 2009 1:53PM
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t_rob
Posts: 6433 - Karma: 11,458
I concur.
Sep 3 2009 1:55PM
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r0dubs
Posts: 1764 - Karma: 19
wow im dumb
Sep 3 2009 1:56PM
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sstricky
Posts: 1876 - Karma: 111
this sounds better... but yea.. how would you do it without taking the derivatives?
Sep 3 2009 2:00PM
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ScuffDUFF
Posts: 2027 - Karma: 143
I just finished calculas finally.   Therefor im done, and not gunna grab my notes to try and find the limits of this equation.   Have fun with Crackulus.
Sep 3 2009 2:06PM
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ORski21
Posts: 423 - Karma: 12
um... i dont think you can. They're easy derivatives though, i didnt use a calculator for any of that problem
Sep 3 2009 2:12PM
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musick13
Posts: 913 - Karma: 44
I see
Sep 3 2009 2:13PM
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ChronicLove
Posts: 84 - Karma: 31
i start calc in four days. this thread just makes me even more depressed about it. thanks guys....
Sep 3 2009 2:21PM
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ScuffDUFF
Posts: 2027 - Karma: 143
i was taught that infinity over infinity would just be equal to 1.  I could be wrong though, i sleep alot.
Sep 3 2009 2:22PM
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ScuffDUFF
Posts: 2027 - Karma: 143

or no its infinity haha fml

Sep 3 2009 2:24PM
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70794Willow
Posts: 334 - Karma: 38
same here.

i hate when shit like that happens and i can't remember things i used to be good at. it's always such a downer for myself
^i guess i'm just expecting too much of myself sometimes...

Sep 3 2009 2:28PM
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ChronicLines
Posts: 951 - Karma: 61
Gotta differentiate everything a couple times, leaving you with -30x+6/6 (as mentioned before).


So yes, the answer is negative infinity.
Sep 3 2009 2:47PM
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70794Willow
Posts: 334 - Karma: 38
k now i'm rememebering after looking at it for a while.
i'm sry to be a nerd, but here are some tips for having to evaluate limits going to infiinty.

compare the highest degree of the funtion. that's the biggest exponent of the variable. in this case it's 3. so, u basically ignore all numbers and variables with an exponent less than that highest degree. now there are only 3 possibilites:

1) if there's a variable with the highest degree on the top and bottom of the function, then your limit is always just the coefficents divided by themselves. (watch out for signs!)
2) if only the variable on top of the function has this highest degree, then it's either negative or positive infinity. again, depending on the sign.
3) if the variable on the bottom has the highest degree of the funtion, the limit is undefined since u can't divide by negative or positive infinty.


hope this made sense. this way i used to be able to rush through the calc homework when we had to evaluate limits at infinity.
Sep 3 2009 2:51PM
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hoodSlayer
Posts: 900 - Karma: 7
for future reference when finding a limit when x goes to +/- infinity just divide my the highest degree in the denominator
Sep 3 2009 2:52PM
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70794Willow
Posts: 334 - Karma: 38
sry for triple post but so in this case you would just look at -5x^3 and see the negative and since -5x^3 is at the top, you know it's negatvie infinity.
problem solved within 5 seconds. :)

i really hope this makes sense to you.
Sep 3 2009 2:54PM
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prontocuts.com
Posts: 5244 - Karma: 42
my motivation for finishing calc is getting a tattoo.

as soon as I finish lin dif alg, im getting a tattoo of my favorite formula to show how bad ass of a nerd I am
Sep 3 2009 2:55PM
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sstricky
Posts: 1876 - Karma: 111
i got the answer.

so you multiply through by 1/x^2 over 1/x^2

then everythign with x under it becomes 0
so you get -5(x)/3 so when u plug in negative infinity it would become positive leaving you with positive infinity
Sep 3 2009 3:04PM
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70794Willow
Posts: 334 - Karma: 38
that's right but dont u have to find the limit when x approaches (postive) infinty.
^ looking at your original post.

but, please do yourself a favor and reaad through my posts. if you get what i'm saying, you will solve problems like these within 5 seconds. no, not kidding.
Sep 3 2009 3:09PM
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Jordan
Posts: 4079 - Karma: 2,122
I am pretty sure

Click for larger image


is the answer.
Sep 3 2009 3:36PM
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dstriebs
Posts: 396 - Karma: 77
wow ap calc is gunna kick my ass
Sep 3 2009 3:40PM
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ashamedkrew
Posts: 1194 - Karma: 424
just started calc on tuesday. it looks weird lookin at math problems on newschoolers though haha
Sep 3 2009 4:26PM
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sstricky
Posts: 1876 - Karma: 111
yea i did. thanks man.

i just talked to my dad whos a math teacher and his exact words were

"the only thing that matters are the x in the numerator with the highest degree and the x in the denominator with the highest degree and take everything else away, and then from there plug in and all that matters from now on are the signs of the constant.
Sep 3 2009 4:40PM
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70794Willow
Posts: 334 - Karma: 38
yeah i'm glad it made sense.

exactly, that's how i learned it. now it's incredibly easy to take limiits approaching infinity.
:)

i would say /thread
Sep 3 2009 5:53PM
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An10a_Eng
Posts: 2505 - Karma: 28
power of x is greater on top so that will dominate the bottom. since its -x^3 its infinity. if the power on the bottom is bigger it goes to zero. if they are the same power it goes to the leading coeffecients (2x/x) goes to 2 as x goes to infinity.
Sep 3 2009 7:03PM
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An10a_Eng
Posts: 2505 - Karma: 28
TAKE IT TO THE CALCULUS CULT!
Sep 3 2009 7:03PM
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mr_shifty
Posts: 316 - Karma: 16
wow i really forgot a lot over the summer
Sep 3 2009 7:25PM
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Uncle.Badness
Posts: 2545 - Karma: 55
This is right.
Sep 3 2009 7:27PM
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00_Koda
Posts: 922 - Karma: 269
shit FML im not even gonna step into that course
Sep 3 2009 8:14PM
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An10a_Eng
Posts: 2505 - Karma: 28
calc 1 is easy bones
Sep 3 2009 8:30PM
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NESblowjob
Posts: 1174 - Karma: 41
no dude its not that hard all once you see that the 3x^2 cancels, you just look to see what dominates as x goes to infinity
Sep 3 2009 8:58PM
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