## Calculus (3rd Edition)

$$\lim_{x\to 1}x^3= 1$$
Let $\epsilon\gt0$ be given. We bound $\left|x^{3}-1\right|$ by factoring the difference of cubes: $$\left|x^{3}-1\right|=\left|\left(x^{2}+x+1\right)(x-1)\right|=|x-1|\left|x^{2}+x+1\right|$$ Let $\delta=\min (1, \xi),$ and assume $|x-1|\lt\delta .$ Since $\delta\lt1,0\lt x\lt 2 .$ Since $x^{2}+x+1$ increases as $x$ increases for $x\gt 0, x^{2}+x+1\lt 7$ for $0\lt x\lt 2,$ and so $$\left|x^{3}-1\right|=|x-1|\left|x^{2}+x+1\right|\lt 7|x-1|\lt 7 \frac{\epsilon}{7}=\epsilon$$ Hence $$\lim_{x\to 1}x^3= 1$$