# Evaluate Derivative Using $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ Definition

Evaluate the derivative of $x^3 - 3x +1$ using the $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ definition to find the tangent of the curve at the point $(2, 3)$.

I already calculated this derivative using $x = a + h$ for the above mentioned definition and this is what I got, for what it's worth:

$$\lim_{h \to 0} \frac{((2+h)^3 - 3(2+h) +1) - 3)}{h}$$

$$=\lim_{h \to 0} \frac{h(9 + 6h)}{h} = 9$$

The problem I encounter when using the $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ definition is that I can't factor the numerator of $\lim_{x \to 2} \frac{x^3 - 3x +1 - 3}{x - 2}$ to get rid of that $(x-2)$ in the denominator.

Can anybody give me a hint as to what trickery I can use to factor that numerator?

• Use polynomial division – rlartiga Jul 30 '14 at 14:10
• $$x^3-3x-2=(x-2)(x^2+2x+1)$$ – rlartiga Jul 30 '14 at 14:11
• I understand now, thanks, rlartiga. – Kermit the Hermit Jul 30 '14 at 15:17

Note that $$x^3 - 3x - 2 = (x - 2)(x^2 + 2x + 1) = (x-2)(x+1)^2$$
If all else fails when you are unable to see the factorization, you can use polynomial long division: Dividing, in this case, by $x-2$ yields the quadratic factor $(x^2 + 2x + 1)$.