#1
I have a homework problem I'm stuck on for my math class.

I need to find:

the limit as x approaches 5 from the + of f(x).

f(x) = e^x /((x-5)^3)

Any help?
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#3
I could have helped you last year.

Unfortunately, everything I learnt at school has leaked out my ears, and I'll be surprised if I get anywhere in life. And guess what? It'll happen to you! Then you'll be praying for death.

Good luck!

DON'T MAKE ME DESTROY YOU!


___________________________________________________


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#6
multiply out the denominator, and change the top to a natural log (ln). then plug in 5 for x
#8
Thanks for those of you that helped me. I got it now
Quote by Sonicxlover
I once told a Metallica fan I liked Megadeth, and he stabbed me 42 times.
#9
Lo hopital's rule? that might help out, mind you, you might have to take the 3 derivative or so
#11
Quote by maggot9779
The answer is:

42

hitch hikers guide to the galaxy... classic

yeah, i hate limits, stupid calc
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#12
Quote by TrUe MeTaL FaN
wut in the hell does that mean ahh im in 10th grade geometry


if your school's curriculum is like mine, you'll be doing that next year in algebra 2
#13
Quote by Samdunhamss
Lo hopital's rule? that might help out, mind you, you might have to take the 3 derivative or so


Do you even know what L'Hopital's rule is used for?
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Art & Lutherie
#17
use l'hopital's rule

when you've got a limit of a fraction where you've got 0 in the denominator you can differentiate both the numerator and denominator and the value will stay the same

so we've got a/b
a= e^x
b = (x-5)^3

and we just keep differentiating them until we get a b != 0 when we put in x=5

a' = e^x
b' = 3*(x-5)^2

a'' = e^x
b'' = 6*(x-5)^1

a''' = e^x
b''' = 6


so limit x->5 for (e^x /((x-5)^3)) = limit x->5 for (e^x/6) = 24.7355265


wikipedia the rule
#18
Quote by seljer
use l'hopital's rule

when you've got a limit of a fraction where you've got 0 in the denominator you can differentiate both the numerator and denominator and the value will stay the same

so we've got a/b
a= e^x
b = (x-5)^3

and we just keep differentiating them until we get a b != 0 when we put in x=5

a' = e^x
b' = 3*(x-5)^2

a'' = e^x
b'' = 6*(x-5)^1

a''' = e^x
b''' = 6


so limit x->5 for (e^x /((x-5)^3)) = limit x->5 for (e^x/6) = 24.7355265


wikipedia the rule


Can't use it. If the question is lim(x->a) of f(a)/g(a) then one condition is:
f(a) = g(a) = 0 or [ lim f(x) = plusminus infinity and lim g(x) = plusminus infinty ]

The answer is infinity.

(x-5) becomes arbitrarily small as x tends to 5 from above. That means that (x-5)^3 becomes even smaller.

We can treat the top e^5 as constant.

Means that it is a constant divided by an arbitrarily small constant that is slightly above 0 (i.e. epsilon). That equals infinity.
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That is part of the reason that the mafia does so much drug trafficing, its so they wont die of hunger because they dont have anything.
#19
Quote by sinisa
Can't use it. If the question is lim(x->a) of f(a)/g(a) then one condition is:
f(a) = g(a) = 0 or [ lim f(x) = plusminus infinity and lim g(x) = plusminus infinty ]

The answer is infinity.

(x-5) becomes arbitrarily small as x tends to 5 from above. That means that (x-5)^3 becomes even smaller.

We can treat the top e^5 as constant.

Means that it is a constant divided by an arbitrarily small constant that is slightly above 0 (i.e. epsilon). That equals infinity.

bah, its been a year since i took that class
#20
Quote by seljer
bah, its been a year since i took that class


tisk tisk

As my old 1st stage maths prof would say (with a thick chinese accent and a beard like pei mey, no shit):
Know your definitions

You studying maths or what?
Quote by bassplayer33333
Sinisa Rules all.


Quote by zbest
That is part of the reason that the mafia does so much drug trafficing, its so they wont die of hunger because they dont have anything.