Prove the seq $\{\frac{x^4-1}{x^4+x-6}\}$ converges and prove using $N-\epsilon$ proof Prove $$\{\dfrac{x^4-1}{x^4+x-6}\}$$ converges and prove using $N-\epsilon$ proof.
I see that this sequence approach $1$.
So,
I want to show:
$$\forall \epsilon >0, \exists N>0, s.t, n>N \to |\dfrac{x^4-1}{x^4+x-6}-1| < \epsilon$$
I started this proof and was trying to find an ideal $N$.
$$|\dfrac{x^4-1}{x^4+x-6}-1| < \epsilon$$ 
$$|\dfrac{x^4-1}{x^4+x-6}-\frac{(x^4+x-6)}{x^4+x-6}| < \epsilon$$
$$|\dfrac{-1-x+6}{x^4+x-6}| = |\dfrac{5-x}{x^4+x-6}| = |\dfrac{x-5}{x^4+x-6}| < \epsilon$$
Now I wanted to do some bounding so I can make the function larger:
$$|\dfrac{x-5}{x^4+x-6}| \leq |\dfrac{x-5}{x^4}| \leq |\dfrac{2x}{x^4}| \leq |\dfrac{2}{x^3}|$$
but this is not the case when I graph it. Where is my bounding wrong?

As you can see, the red graph is actually bigger then the blue graph. Why?

Note that for $n\ge 2$, $n^4+n-6\ge n^4/2$.  Hence we can write 
$$\left|\frac{n^4-1}{n^4+n-6}-1\right|=\left|\frac{n-5}{n^4+n-6}\right|\le \frac{n}{n^4/2}=\frac{2}{n^3}<\epsilon$$
whenever $n>(2/\epsilon)^{1/3}$.
