Common Calculus Mistakes #5 #Shorts #calculus #mistakes #math #maths #mathematics

[carultch]

They are related to each other. Here's a limit proof that the first method isn't entirely wrong, as long as you don't directly divide by zero.

Start with:
integral x^(-1 + h) dx = 1/h * x^h + C
where h is a number very close to zero

Let C = -1/h:
1/h * x^h - 1/h
(x^h - 1)/h

Replace x^h with e^(ln(x)*h):
(e^(ln(x)*h) - 1)/h

Multiply by ln(x)/ln(x), and take the numerator out in front of the limit:
ln(x)*(e^(ln(x)*h) - 1)/(h*ln(x))

Let H = ln(x)*h:
ln(x)*(e^H - 1)/H


Take the limit as H goes to zero, which is the same as lowercase h going to zero.
ln(x) * limit H->0 of (e^H - 1)/H

Recognize the limit of (e^H - 1)/H, as the definition of the derivative of the exponential function at an input of zero, which is equal to 1. Thus, the limit evaluates to 1.

This means our result is:
integral 1/x dx = ln(|x|) + C

The nature of complex log introduces the absolute value signs, when setting it up for the real number log to work for the integral 1/x on all possible reals..