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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.

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

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  • $\begingroup$ where exactly does 1 come from when x -> -1? $\endgroup$ – Sander Sep 13 '17 at 17:13
  • $\begingroup$ @JoakimHauger Sorry, my mistake. Now it is correct. $\endgroup$ – Robert Z Sep 13 '17 at 17:15
  • $\begingroup$ okei thanks, im still not 100% where 1/2≤|x|≤2 comes from $\endgroup$ – Sander Sep 13 '17 at 17:21
  • $\begingroup$ What you really need is that $\delta<1$ then $|x+1|<\delta$ implies that $0<1-\delta <|x|<1+\delta$. $\endgroup$ – Robert Z Sep 13 '17 at 17:23
  • $\begingroup$ okei, thanks. i think i got it now. $\endgroup$ – Sander Sep 13 '17 at 17:29

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