Expert: Socrates - 10/22/2007

Question
hi, am having a little trouble understanding some of the ways to solve limits
by differentiation. this particular problem, the teacher did but didn't explain.
could you please explain it to me?

lim ( 1+1/x)^ 1/x
x->0+0
with the answer being e.

Answer
I think you wanted lim (1+1/x)^x , since the limit you have asked for is clearly 1 , without using L'Hospital's rule.

To find lim(1+1/x)^x , we can find lim ln(1+1/x)^x first. This will be easier , since it brings the exponent down as a factor. We have ln(1+1/x)^x = (x)ln(1+1/x) =
ln(1+1/x)/(1/x) . Note that the numerator and denominator both have limit 0 when x goes to positive infinity. When a fraction has zero as the limit in the numerator and denominator , you can use L'Hospital's rule to find the limit of the fraction. This just means to replace the numerator and denominator by their derivatives and then finding the limit of the new expression. So in your problem , you find [ln(1+1/x)]' = -1/(x^2+x) and
[1/x]' = -1/x^2 . Then [-1/(x^2+x)]/[-1/x^2] = x/(x+1).
You know that lim x/(x+1) = 1 as x goes to positive infinity , so you now have lim ln(1+1/x)^x = 1 . Since
ln(1+1/x)^x is going to 1 , (1+1/x)^x must be going to e .
lim (1+x)^1/x = e