Prove the seq $\{\frac{x^4-1}{x^4+x-6}\}$ converges and prove using $N-\epsilon$ proof

Question

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?

Answer 1

$|\dfrac{x-5}{x^4+x-6}| < \epsilon$

Now start bounding $N$

Suppose $N> 6$

$|x-5| < N\\ |x^4+x-6| > N^4$

$|\dfrac{x-5}{x^4+x-6}| < \frac 1{N^3}$

$\frac 1{N^3} = \epsilon\\ N = \max (6,\frac {1}{\sqrt[3]{\epsilon}})$

Answer 2

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 $n>(2/\epsilon)^{1/3}$.