How to calculate the limit of this function (Advanced Calculus)?

Question

I want to ask how to calculate this lim

lim (1 - 2/x) ^x when x --> infinity

Thx a lot
(dont use graphing method, I have tried L'Hosiptal but havent figured out)

And the graphing method give the answer .135335 so dont be silly when say it's 1 . 1^infinity is not 1

Answer 1

e^{-2}

you can NOT use L'Hopital's rule in this case,
and they way to prove it is by using the substitution
y=- x/2
then
lim x-> infinity (1 - 2/x) ^x = lim y-> infinity (1 +1/y) ^{-2y}
= lim y-> infinity [ (1 +1/y) ^{y}]^{-2}
= [ lim y-> infinity (1 +1/y) ^{y} ]^{-2}
=(e)^{-2}

Answer 2

When x = infinity, (1 - 2/x) = 1
1 to any power, even infinity = 1
So, the limit is 1.

Answer 3

be conscious that 4x - x^2 = x(4 - x) = x(2 + sqrt(x))(2 - sqrt(x)) now sub in to furnish: lim (2 - sqrt(x)) / [x(2 + sqrt(x))(2 - sqrt(x)) (2 - sqrt(x)) / (2 - sqrt(x)) cancels to a million as long as x =/= 4, and is a detachable discontinuity you will be able to desire to replace in 4 for x in what's left to locate the shrink a million / [x(2 + sqrt(x)] as x ==> 4 is a million / [4(2 + 2)] = a million/sixteen shrink as x ==> 4 = a million/sixteen as quickly as you initially substitue 4 for x, you get 0 / 0, it quite is indeterminate, telling you which you would be waiting to look for factors of 0 to cancel..

Answer 4

thats right