Use the $\epsilon$-$\delta$ definition of a limit to prove $\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$ $\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$
You should start by writing $\frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6} + 1$ in the way $\left ( x-3 \right )g\left ( x \right )$
(a) Determine $g(x)$
(b) Could we choose $\delta = \min\lbrace 1, \frac{\varepsilon }{n} \rbrace$ for some $n$? Why?
(c) If we choose $\delta = \min\lbrace \frac{1}{4}, \frac{\varepsilon }{m} \rbrace$ , what is the smallest integer $m$ that we could use?

I managed to answer the part (a): By making the sum of the fraction with +1, and factoring the numerator, I get the following:
$\frac{\left ( x-3 \right )\left ( x^{3} - x^{2} - 2x - 4 \right )}{x^{4} - 4x^{3} + x^{2} + x + 6}$
where
$g\left ( x \right ) = \frac{\left ( x^{3} - x^{2} - 2x - 4 \right )}{x^{4} - 4x^{3} + x^{2} + x + 6}$
But my problem is when developing the part (b).
In the part (b) I am using the formal limit definition, but I am not managing to reach a result that allows me to find the answer.
I need help to solve this exercise, please
 A: This is a bit complicated question. You need to find a suitable bound for the rational function $g(x) $ for values of $x$ in a certain neighborhood of $3$. In formal terms we need two positive numbers $h, K$ such that $|g(x) |<K$ whenever $|x-3|<h$.
This clearly requires an analysis of numerator and denominator of $g$. Note that the function can not be bounded if the denominator of $g$ vanishes. Since it does not vanish at $3$ we have a guarantee by continuity of polynomials that the denominator will not vanish in some neighborhood of $3$.
Finding such a neighborhood (ie finding $h$ described above) is possible using the limit of denominator $q(x) =x^4-4x^3+x^2+x+6$ as $x\to 3$. But before we do that let's observe that $q(3)<0,q(4)>0$ and hence $q$ vanishes somewhere between $3,4$ and thus $h$ must be less than $1$.
Thus for part b) the answer is no. You can't expect a $\delta$ like $\min(1,\dots)$ to work. You should observe that $h$ described above forms the first argument of $\min$ in expression for $\delta$.
Part c) is about showing that we can take $h=1/4$. Let us now prove that $q(x)\to - 9$ as $x\to 3$. Clearly $$|q(x) +9|=|x-3||x^3-x^2-2x+5|$$ and $$|x^3-x^2-2x+5|=|x(x-2)(x+1)+5|$$ If $|x-3|<1/2$ then $5/2<x<7/2$ and hence $$|x(x-2)(x+1)+5|<(7/2)(3/2)(9/2)+5<32$$ and thus for any $\epsilon>0$ we have $$|q(x) +9|<\epsilon$$ if $|x-3|<\min(1/2,\epsilon /32)$. Using $\epsilon =8$ we can see that $$|q(x) +9|<8$$ if $|x-3|<1/4$. Thus $|q(x) |>1$ if $|x-3|<1/4$.
Next we deal with numerator $p(x) =x^3-x^2-2x-4$ of $g(x) $. Clearly if $|x-3|<1/4$ then $2.5<x<3.5$ and $$|p(x) |=|x(x-2)(x+1)-4|<28$$ Thus it follows that if $|x-3|<1/4$ then $$|g(x) |=|p(x) |/|q(x) |<28$$
It should now be clear that $$|(x-3)g(x)|<28|x-3|$$ and this is less than $\epsilon$ if $|x-3|<\min(1/4,\epsilon/28)$. This proves the desired limit in question.
It does not however make sense to find the smallest positive integer which can replace $28$ above. Finding such numbers becomes an algebraic problem related to solution of inequalities and is really not in the spirit of epsilon-delta definition of limit.
