lim g(x) .
x=>oo f(x)
2006-07-08 14:01:49 · 10 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
Let f(x) = e^3x - 1 and g(x) = e^4x + 2
Evaluate:
lim .
x=>oo
g(x)
f(x)
2006-07-08 14:09:24 · update #1
L'hopital's rule
thanks
2006-07-08 14:10:09 · update #2
Is that the limit as x goes to infinity of g(x) divided by f(x)?
(e^4x + 2)/(e^3x - 1)
This goes to infinity/infinity.
Use L'hopital's rule
4e^4x/(3e^3x)
= (4/3)e^x
which goes to infinity.
2006-07-08 14:06:59 · answer #1 · answered by MsMath 7 · 3⤊ 0⤋
Lim ( e^3 x - 1) / (e^4 x +2)
x ââ
2006-07-09 05:20:54 · answer #2 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
I assume you mean (xââ)lim g(x)/f(x). In which case, the answer is â. If this is not obvious, consider the substitution y=e^x. Then your problem becomes (yââ)lim (y^4+2)/(y^3-1), and that should be obvious.
2006-07-08 21:10:44 · answer #3 · answered by Pascal 7 · 0⤊ 0⤋
take e^4x (1+2e^-4x) 1+2e^-4x
----------------------- = --------------
e^3x -1 e^-x - e^-4x
substitute limit ie. x=infinity
as e^-x -> 0 as x tends to infinit , the denominator becomes 0
and numerator is 1-0=0
the answer. therefore is 1/0 = 00
2006-07-08 22:38:07 · answer #4 · answered by Srikanth 2 · 0⤊ 0⤋
Both functions grow very fast to infinity when x goes to infinity.
2006-07-09 01:36:44 · answer #5 · answered by Thermo 6 · 0⤊ 0⤋
Ooooo, stuff w/ infinity. Sorry, I just finished Advanced Algebra. lol.
2006-07-08 22:06:29 · answer #6 · answered by agfreak90 4 · 0⤊ 0⤋
Um, what are you asking for, it didn't come through nicely when you posted it...
2006-07-08 21:05:51 · answer #7 · answered by aanusze1 3 · 0⤊ 0⤋
thats ridiculous
2006-07-08 21:05:11 · answer #8 · answered by Ray 4 · 0⤊ 0⤋
interesting...
2006-07-08 21:05:06 · answer #9 · answered by Anonymous · 0⤊ 0⤋
no
2006-07-08 21:03:15 · answer #10 · answered by Shawn B 2 · 0⤊ 0⤋