Slope of the tangent line using the limit definition

Substitute into the difference quotient: $\frac{f(2+h) - f(2)}{h} = \frac{(2+h)^3 - 2^3}{h}.$ Expanding $(2+h)^3 = 8 + 12h + 6h^2 + h^3$, so the numerator becomes $8 + 12h + 6h^2 + h^3 - 8 = 12h + 6h^2 + h^3$.

Divide every term by $h$ to get $12 + 6h + h^2$, since $h \ne 0$ in the limit process. Taking the limit as $h \to 0$ gives $12 + 0 + 0 = 12$. Therefore $f'(2) = 12$, which is the slope of the tangent line to $y = x^3$ at $x = 2$.