Use the $\varepsilon -\delta$ definition of limit to show that $\displaystyle\lim_{x\to 1} (x^3+x+1)=3$
Here's my attempt:
When $|x-1|<\delta$ we have $$|x^3+x+1-3|=|x-1||x^2+x+2|<\delta |x^2+x+2|$$
Assuming $|x-1|<1$, we have that $-1<x-1<1$, which implies that $|x|<2$ and $|x^2|<4$.
Now, by the triangle inequality we get $|x^2+x+2|\leq |x^2|+|x|+|2|<4+2+2=8.$
Let $\delta =$min$\left(1,\frac{\varepsilon}{8}\right)$, then $$|x^3+x+1-3|<\delta|x^2+x+2|$$ $$<8\delta$$ $$\leq 8\frac{\varepsilon}{8}=\varepsilon$$
Is this correct?