The linear approximation to f(x) = ln(x) at x = 1 is
L(x) = f(1) + f'(1) (x - 1)
We have f(1) = ln(1) = 0, and the derivative of f is
f'(x) = 1/x
so that f'(1) = 1/1 = 1. Then
L(x) = 0 + 1 (x - 1)
L(x) = x - 1
We then approximate ln(1.25) by
ln(1.25) ≈ L(1.25) = 1.25 - 1 = 0.25
