can someone please help with this: limit of x^(1/x) as x approaches infinity
S spaceface New member Joined Sep 6, 2006 Messages 1 Sep 6, 2006 #1 can someone please help with this: limit of x^(1/x) as x approaches infinity
R ratzlaff New member Joined Sep 6, 2006 Messages 1 Sep 6, 2006 #2 This limit is an indeterminate form. In this case it is the (infinity)^0 form. You have to use l'Hopital's Rule to solve this. 3 steps to solve this 1. let y=f(x)^(g(x)) 2. take the natural log of both sides to get lny=g(x)ln(f(x)) using the properties of logarithms. then find the lim of lny. 3. lim of y as x approaches a = lim of e^lny as x approaches a = e^lim of lny as x approaches a. so basically when you are done with step two, then put that answer as the power of e. the bulk of this problem is finding the limit of ln(y) if you are not familiar with l'Hopital's Rule it says that lim of f(x)/g(x) = lim of f'(x)/g'(x) so to find lim of g(x)ln(f(x)) you have to change it to look like this lim of ln(f(x))/(1/g(x))
This limit is an indeterminate form. In this case it is the (infinity)^0 form. You have to use l'Hopital's Rule to solve this. 3 steps to solve this 1. let y=f(x)^(g(x)) 2. take the natural log of both sides to get lny=g(x)ln(f(x)) using the properties of logarithms. then find the lim of lny. 3. lim of y as x approaches a = lim of e^lny as x approaches a = e^lim of lny as x approaches a. so basically when you are done with step two, then put that answer as the power of e. the bulk of this problem is finding the limit of ln(y) if you are not familiar with l'Hopital's Rule it says that lim of f(x)/g(x) = lim of f'(x)/g'(x) so to find lim of g(x)ln(f(x)) you have to change it to look like this lim of ln(f(x))/(1/g(x))
