I prefer to do all the algebra BEFORE taking a limit.
[MATH]f(x) = x^n,\ n \text { is an integer} > 1.[/MATH]
[MATH]\Delta x \ne 0.[/MATH]
[MATH]f(x + \Delta x) = (x + \Delta x)^n = \sum_{j=0}^n \dbinom{n}{j} x^{(n - j)} \Delta x^j = x^n + nx^{(n-1)} \Delta x + \Delta x^2 \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ).[/MATH]
[MATH]\therefore f(x + \Delta x) - f(x) = nx^{(n-1)} \Delta x + \Delta x^2 \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right ) \implies[/MATH]
[MATH]\dfrac{f(x + \Delta x) - f(x)}{\Delta x} = nx^{(n-1)} + \Delta x \left ( \sum_{j=2}^n \dbinom{n}{j} x^{(n-j)} \Delta x^{(j-2)} \right )
[/MATH]
Now take the limit. As delta x goes toward zero, is it not intuitive that the term multiplied by delta x also goes toward zero leaving the other term unaffected?