prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit 
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this

Let $\varepsilon > 0$. We have
$$\left|\frac{x^3-1}{x-1} - 3\right| = \left|\frac{x^3-3x+2}{x-1}\right| = \left|x^2+x-2\right| = |x-1||x+2| \le |x-1|(|x-1| + 3)$$
Solving the quadratic inequation $y^2+3y - \varepsilon < 0$ gives $y \in \left\langle \frac{-3-\sqrt{9+4\varepsilon}}2, \frac{-3+\sqrt{9+4\varepsilon}}2\right\rangle$.
Therefore if we pick $\delta = \frac{-3+\sqrt{9+4\varepsilon}}2$, then $|x-1| < \delta$ implies 
$$\left|\frac{x^3-1}{x-1} - 3\right| \le |x-1|^2 + 3|x-1| < \varepsilon$$
We conclude $\lim_{x\to 1} \frac{x^3-1}{x-1} = 3$.
