What is the lim as x approches infinty of (x/(x+1))^x

Cal exam in 6 hours. its 3 am...

Let L = lim[x→∞] (x/(x+1))^x
ln(L) = lim[x→∞] ln((x/(x+1))^x)
ln(L) = lim[x→∞] x ln(x/(x+1))

Let u = x/(x+1)
Then x = u/(1−u)
As x→∞, u = x/(x+1)→1

ln(L) = lim[u→1] u/(1−u) ln(u)
ln(L) = lim[u→1] u ln(u) / (1−u) = 0/0 -----> Use L'Hopital's Rule
ln(L) = lim[u→1] (u*1/u + 1*ln(u)) / −1
ln(L) = lim[u→1] (1 + ln(u)) / −1
ln(L) = (1 + ln(1)) / −1 = (1+0) / −1
ln(L) = −1
L = e^(−1)
L = lim[x→∞] (x/(x+1))^x = 1/e

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@Mike, when x gets very large, then (x/(x+1))^x gets close to 0.367879441 (1/e), not 0

x = 1 -----> (1/2)^1 = 0.5
x = 10 ----> (10/11)^10 = 0.385543289
x = 100 ----> (100/101)^100 = 0.369711212
x = 1000 ----> (1000/1001)^1000 = 0.368063304
x = 10000 ----> (10000/10001)^10000 = 0.367897834
x = 100000 ----> (100000/100001)^100000 = 0.367881281
x = 1000000 ----> (1000000/1000001)^1000000 = 0.367879625
http://www.google.ca/#hl=en&sclient=psy-…

Sorry the answer is zero, I didnt see the exponent part of the problem....

The best way to figure an infinite limit out is to plug in 1 for x, then go big and plug something like one million in. The equation gets closer and closer to zero