TITLE: $\epsilon-\delta$ limit proof
QUESTION [5 upvotes]: Using the epsilon delta definition of limits prove:
  $$\lim\limits_{x \to -1} \frac{x^4+x+1}{x^3}=-1.$$  

I have managed to get 
$$\left|{\frac{x^4+x+1}{x^3}}+1\right| =  \frac{\vert x+1\vert^2\vert x^2-x+1 \vert}{|x|^3}$$ 
which is a step closer I think since I have 
the factor $(x+1)$ which I can control. And I can also limit the other factor in the numerator.
But the $x^3$ in the denominator is my problem because if I limit $(x+1)$ it seems to grow. 
I not sure what to do with it.

REPLY [6 votes]: Hint. Note that if $x\to -1$, we can assume that $1/2\leq |x|\leq 2$ (take $\delta\leq 1/2$) which implies 
$$0<\frac{1}{|x|^3}\leq  8\quad \mbox{and}\quad |x^2-x+1|\leq |x|^2+|x|+1\leq 7.$$