#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?

I need to find:

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

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

Any help?

#2

The answer is:

42

42

#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!

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!

#4

This is the pit. No one can or will help you.

#5

well, the answer is infinity because you'd end up with e^x/a really small number iuf you chose like 5.0001 as x.

#6

multiply out the denominator, and change the top to a natural log (ln). then plug in 5 for x

#7

wut in the hell does that mean ahh im in 10th grade geometry

#8

Thanks for those of you that helped me. I got it now

#9

Lo hopital's rule? that might help out, mind you, you might have to take the 3 derivative or so

#10

Thanks for those of you that helped me. I got it now

It was 42 wasn't it?

#11

The answer is:

42

hitch hikers guide to the galaxy... classic

yeah, i hate limits, stupid calc

#12

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

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?

#14

you need 0/0 or infinity/infinity for l'hopital bud, and that doesn't really happen here

#15

My brain just exploded

#16

I think it's 3.

Yeah, I'm going with 3.

Yeah, I'm going with 3.

#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

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

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.

#19

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

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?