Math Gurus- help on indeterminate limits plz :(

[magomago]

i've done these before...hah but i got a 2 on the B/C test

so here i am in college math repeating this. Now i usually don't attend lecture b/c its pointless (49/50 on first midterm) since i really do know the stuff...but rite now i wanna rip out my hair....

I can't solve this



lim (1+9/x)^(x/10)
x-->INFINITY


I can't solve that...i noe that it is e^9/10 b/c of my ti89, but i would like to see the steps. I also know the definition of e is the same thing, but the 9 and 10 replaced with a 1

but i can't get this...

Cany anyone help me out? I'm searching sites and i found this http://bradley.bradley.edu/~delgado/121/Lhopital.pdf and on page 5 in the middle it shows the steps to defining "e", but I make no sense of the last few stesps where it goes from "the lim as x approaches inf of 2x/2x" to the next step of "therefore the limit must be 2"...

if anyone can help i'd really appreacite it- thanks

 

[TerryMathews]

well, 9/infinity is 0. 1+0 = 1. infinity/10 is infinity. 1^infinity is indeterminate, so apply L'Hopital's rule.

That should be all you need to do it yourself.

 

[magomago]

lol i've been doing that all nite...if i had a place to upload a pic i'd do the work and show it because i just can't get it regardless...

could you type it out? If you did I'd love you to death and send you a postcard thanking you personally or something (i'd have to do that next weekend, midterms all this week)

 

[jmcoreymv]

You want to take the ln of it in order to bring the x/10 down. So you get (x/10) ln (1+ 9/x) which is indeterminant inf * 0 so you rearrange it to get 0/0 by flipping the x/10 to 10/x and dividing with it. So you get ln(1+9/x) / (10/x). Use Lhospitals rule, so you get -9/((1+9/x)(x^2)) and the bottom is -10/x^2 so the x^2s cancel and you get -9/((-10)( 1+9/x)) when you take that to infinity, 9/inf = 0 so you end up with 9/10's. And since you took the ln at the beginning, you need to raise as the power of e. So you get e ^ (9/10)'s.

 

[bonkers325]

you're never going use this in real life so just look in the back of the book for answers. l'hopital's rule btw. turn it into 0/0 or 1^infinity or 0*infinity or (infinity - infinity)... i forget about the last one. in this case 1^infinity would work.

 

[magomago]

thanks guy with the devil icon!!!!!!!!!!!


what i forgot was i still had LN(Y)!!!

I'm sitting here thinking "WTF I KEEP GETTING 9/10" thinking "how did they get e in there"


as for never using it- its called "showing work" to a professor who gives you a grade

and considering i expect to get a C in physics...i need an A in math to balance that C

 

[Ameesh]

L'Hopitals!

 

[TerryMathews]

You owe me big time.

ln y = x/10 ln (1+9/x)
lim x-> inf ln y = inf * 0

L.H.

(ln(1+9/x))/(10/x)

(-9/(x^2+9x)) * (-x^2/10) = (9x^2/(10x^2+90x)) = (9x/(10x+90))

L.H.

9/10

ln y = 9/10
e ln y = 9/10 e
y = e^(9/10)

 

[Goosemaster]

Originally posted by: TerryMathews
You owe me big time.

ln y = x/10 ln (1+9/x)
lim x-> inf ln y = inf * 0

L.H.

(ln(1+9/x))/(10/x)

(-9/(x^2+9x)) * (-x^2/10) = (9x^2/(10x^2+90x)) = (9x/(10x+90))

L.H.

9/10

ln y = 9/10
e ln y = 9/10 e
y = e^(9/10)

The fact that you have to take the natural log before anything always gets everything

remember this?


lim (1 + 1/x)^(x)
x-->INFINITY


=e

 

[Goosemaster]

 

[Goosemaster]

 

[Goosemaster]

 

[Goosemaster]

 

[Goosemaster]

 

[Goosemaster]

Sorry about that, the message board has been giving me errors lately

 

[magomago]

wow..... double, triple, quadruple, quintelpal, sextepal....

SEPTEPAL post

haha that is crazy...the most i've ever done is a double post