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?
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).
I forgot about that expansion method. Will have to look it up. Thanks!
To find the limit of the function (ln(1+x))/x as x approaches 0, you can use a different approach called the Maclaurin series expansion.
The Maclaurin series expansion of ln(1+x) is given by:
ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...
So, substituting this expansion into the initial function:
(ln(1+x))/x = (x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...) / x
Now, let's simplify this expression:
(ln(1+x))/x = 1 - x/2 + x^2/3 - x^3/4 + ...
As x approaches 0, all the terms containing positive powers of x (x^2, x^3, etc.) will become insignificant compared to the first term (1) and the term proportional to x (-x/2).
Therefore, by considering only the first two terms:
(ln(1+x))/x ≈ 1 - x/2
Now, let's take the limit as x approaches 0:
lim(x→0) (ln(1+x))/x = lim(x→0) (1 - x/2) = 1 - 0/2 = 1
So, the limit of (ln(1+x))/x, as x approaches 0, is 1.
No need for L'Hôpital's rule in this case!