Derivative via the limit definition

Compute $f(x+h) = 3(x+h)^2 - 5(x+h) + 2 = 3x^2 + 6xh + 3h^2 - 5x - 5h + 2$. Subtracting $f(x) = 3x^2 - 5x + 2$ gives $f(x+h) - f(x) = 6xh + 3h^2 - 5h$, since the $3x^2$, $-5x$, and $2$ terms cancel.

Divide by $h$: $\frac{f(x+h)-f(x)}{h} = \frac{6xh + 3h^2 - 5h}{h} = 6x + 3h - 5$ after factoring out $h$ from the numerator and canceling. Taking the limit as $h \to 0$ removes the $3h$ term, leaving $f'(x) = 6x - 5$.