lim ................. (e^3x - 1)/(e^x - 1)
x=>0
thank you
2006-09-24 04:29:36 · 6 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
(e^3x - 1) is also (e^x-1)(e^2x+e^x+1) (from a^3-b^3 = (a-b)(a^2+ab+b^2)
so the ration becomes just (e^2x+e^x+1) and your limit is therefore
e^0+e^0+1 = 3
2006-09-24 04:36:50 · answer #1 · answered by sebourban 4 · 1⤊ 0⤋
Put y = e^x. Then, y => 1 as x => 0.
Thus, the problem reduces to
lim (y^3 - 1) / (y - 1) ==>
lim (y^2 + y + 1) ==> 3
y=>1
Here, we have used the formula y^3 - 1 = (y - 1) (y^2 + y + 1).
2006-09-24 11:44:25 · answer #2 · answered by Problem Child 2 · 1⤊ 0⤋
Answer 3
it is in the form 0/0
take derivative
3e^3x / e^x
3/1=3
2006-09-24 11:34:23 · answer #3 · answered by iyiogrenci 6 · 1⤊ 0⤋
lim x>0(e^x-1)(e^2x+e^x+1)/(e^x-1)
=3
2006-09-24 11:44:38 · answer #4 · answered by raj 7 · 1⤊ 0⤋
Done, and I think I got it right too. What, you want me to give you the answer? Sorry, not worth it, and I'll get the two points anyway. Heh heh heh.
2006-09-24 11:31:00 · answer #5 · answered by Anonymous · 0⤊ 3⤋
?
2006-09-24 11:30:37 · answer #6 · answered by Anonymous · 0⤊ 2⤋