Posted by Heather on .
Find the limit as x approaches infinity of (1(1/(2x)))^(2x).

Calculus 2 
Count Iblis,
f(x)= [11/(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. 
Calculus 2 
Steve,
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 
Calculus 2 
Count Iblis,
I see, I forgot a minus sign.
f(x)= [11/(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. 
Calculus 2 
Heather,
thank you! we are supossed to use Lopital's rule though. Is there a different way to do it using Lopital's?

Calculus 2 
Count Iblis,
If you have to use L'HÃ´pital, 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[11/(2x)] / (1/(2x))
This is then a 0/0 case. If you put
1/(2x) = u then this becomes:
Log[1u]/u
and you need to compute the limit of u to zero. 
Calculus 2 
Heather,
thank you so much!