Question
i am confused about what approach to take for (limit as x goes to 0) of (ln(1+x))/x
if i had x going to infinity, i could do l'hospitals, but here?
if i had x going to infinity, i could do l'hospitals, but here?
Answers
drwls
1n(1+x) has Taylor series expansion
x - x^2/2 + x^3/3 + ...
Divide that by x and you get
1 - x/2 + x/3 + ...
The limit as x-> 0 is 1
You could also use L'Hopital's rule for x-> 0. The same rule applies at any singularity (0/0 or infinity/infinity).
x - x^2/2 + x^3/3 + ...
Divide that by x and you get
1 - x/2 + x/3 + ...
The limit as x-> 0 is 1
You could also use L'Hopital's rule for x-> 0. The same rule applies at any singularity (0/0 or infinity/infinity).
Kay
I forgot about that expansion method. Will have to look it up. Thanks!