Given f(x) = ln(x)
To find P, we need to evaluate the sum of f'(0) + f'(3) + f'(6) + ... + f'(2019).
We know that the derivative of ln(x) is 1/x, so we can write:
f'(x) = 1/x
To find f'(0), we need to use the limit definition of the derivative:
f'(0) = lim (h → 0) [ ln(h) - ln(0) ] / h
Since ln(0) is undefined, we can rewrite this expression as:
f'(0) = lim (h → 0) [ ln(h) ] / h
Using L'Hôpital's rule, we can differentiate the numerator and denominator separately:
f'(0) = lim (h → 0) [ 1/h ] / 1
f'(0) = lim (h → 0) [ 1/h ] = ∞
Note that f'(0) is undefined, since ln(0) is undefined, but the limit approaches infinity.
Similarly, we can find f'(3), f'(6), and so on:
f'(3) = 1/3
f'(6) = 1/6
...
f'(2019) = 1/2019
Now we can substitute these values into the sum:
P = f'(0) + f'(3) + f'(6) + ... + f'(2019)
P = ∞ + 1/3 + 1/6 + ... + 1/2019
The sum on the right-hand side is a finite sum of the form:
1/3 + 1/6 + ... + 1/2019 = ∑(1/3n) where n goes from 1 to 673
This is a telescoping sum, meaning that the terms cancel each other out:
∑(1/3n) = (1/3) + (1/6 + 1/9) + (1/12 + 1/15 + 1/18) + ... + (1/2016 + 1/2019)
= (1/3) + (1/3) + (1/3) + ... + (1/3)
= 673/3
Therefore, we have:
P = ∞ + 673/3
Since f'(0) is undefined, P is also undefined.