How do you find the limit of ((2x-3)/(2x+5))^(2x+1) as x goes to infinity?

Question

I know the answer is e^(-8) but I don't know how to get to that.

quit posting if you haven't got a clue...I KNOW i'm supposed to use L'Hospital's Rule but I don't understand how (i have a book in front of me so don't give me examples on how to use it)

anybody got any enlightenment on what to do once you got
ln y = lim [ 2xln(2x-3) - ln(2x-3) - 2xln(2x+5) + ln(2x+5) ]

you can't use l'hospital's rule from this step so what else can you do? any enlightenment?

Answer 1

(2x-3/2x+5)=(2x+5-8)/2x+5 =(2x+5/2x+5) + ((-8)/2x+5) = 1+ ((-8)/2x+5)

You have limit of (1+ (-8)/2x+5)^exp where the exponent exp=((2x+5)/-8)(2x+1)(-8)/(2x+5)
basically you bulid up the limit: (1+1/n)^n which is e for x goes to infinity
The limit above will be = e^lim (2x+1)(-8)/(2x+5) so it is e^(-8)

Answer 2

First, take set the limit equal to y. Then take the ln of both sides. Expand the logarithm and change it to a form that fits La-Hopital's condition. Then, use his rule and you'll find that lny=-8. Therefore,
y=e^-8 and since y also equals the limit, the limit is e^-8.

Answer 3

You replace or plug in the infinity in the X and resolve there are some rules which I don't remember its been a while since I've studied limits, its like solving a problem first solve on top then bottom divide if possible etc

Answer 4

lm (2x-3)/(2x+5) x goes to infinity is 1
so the expression is 1 as x goes to infinity

Answer 5

use L'Hospital's rule (it's been around at least since 1696, so you'd think you'd know it by now)

Answer 6

lol is this foreal? just use a graphing calculator lol

Answer 7

sorry dont know....... IM NOT THE SMARTEST PERSON IN THE WORLD!!!

Answer 8

not sure

Answer 9

okay....