Find the limit as x approaches infinity of (1-(1/(2x)))^(2x).
f(x)= [1-1/(2x)]^(2 x)
Log[f(x)] = 2x Log[1-/(2x)] =
2 x [1/(2x) + O(1/x^2)] =
1 + O(1/x)
The limit for x to infinity of
Log[f(x)] is thus 1. The limit for x to infinity of f(x) is thus e, because the logarithmic function is continuous there.
Hmmm. Let u = 1/2x and you have
(1+u)^(-1/u) = 1 / (1+u)^(1/u)
as x->infinity, u->0, and we have
1/e
I see, I forgot a minus sign.
f(x)= [1-1/(2x)]^(2 x)
Log[f(x)] = 2x Log[1-/(2x)] =
2 x [-1/(2x) + O(1/x^2)] =
-1 + O(1/x)
So, you get exp(-1), using the fact that log is continuous there.
thank you! we are supossed to use Lopital's rule though. Is there a different way to do it using Lopital's?
If you have to use L'Hôpital, then that's close to how I did it above. After taking logarithms, you have:
Log[f(x)] = 2x Log[1-/(2x)]
You can write this as:
Log[1-1/(2x)] / (1/(2x))
This is then a 0/0 case. If you put
1/(2x) = u then this becomes:
Log[1-u]/u
and you need to compute the limit of u to zero.
thank you so much!
To find the limit as x approaches infinity of the expression (1 - 1/(2x))^(2x), we can use the limit properties and the concept of exponential functions.
Step 1: Simplify the expression inside the parentheses.
1 - 1/(2x) = (2x - 1)/(2x)
Step 2: Rewrite the expression as a power of e.
(2x - 1)/(2x) = [(2x - 1)/2x]^2x
Step 3: The expression [(2x - 1)/2x] is of the form (1 - p/n)^n, which is a specific form as n approaches infinity, known as the exponential form.
Step 4: The exponential form (1 - p/n)^n approaches e^(-p) as n approaches infinity. In this case, p = 1 and n = 2x.
Step 5: Therefore, the expression [(2x - 1)/2x]^2x approaches e^(-1) as x approaches infinity.
Step 6: The final limit is e^(-1), which is approximately 0.3679.