I am trying to show that the given limit exists by use of the epsilon-delta definition. Below is my working thus far:
Let $f(x) = \frac{1}{x^3 + 1}$. If $\lim_{x \to \infty} = \frac{8}{9}$, then forany $\epsilon > 0$, there should exist a $\delta > 0$ such that if $|x - \frac{1}{2}| < \delta$,then $|f(x) - \frac{8}{9}| < \epsilon$.
We have\begin{align*} \left|f(x) - \frac{8}{9}\right| &= \left|\frac{1}{x^3 + 1} - \frac{8}{9}\right| \\&= \left|\frac{9 - 8(x^3 + 1)}{9(x^3 + 1)}\right| \\&= \left|\frac{1 - 8x^3}{9(x^3 + 1)}\right| \\&= \left|\frac{-(2x - 1)(4x^2 + 2x + 1)}{9(x^3 + 1)}\right| \\&= \left|\frac{(2x - 1)(4x^2 + 2x + 1)}{9(x^3 + 1)}\right| \\&= \left|\frac{2(x - \frac{1}{2})(4x^2 + 2x + 1)}{9(x^3 + 1)}\right| \\&= \frac{2}{9}\left|x - \frac{1}{2}\right| \left|\frac{4x^2 + 2x + 1}{x^3 + 1}\right|\end{align*}That is, I have managed to attain my "delta" expression: $|x - \frac{1}{2}|$ in my workings. However, I am unable to continue from here - I no longer know how to continue manipulating\begin{equation}\left|\frac{4x^2 + 2x + 1}{x^3 + 1}\right|\end{equation}to produce anything meaningful. I tried to separate this expression into a sum of modulo fractions using triangle inequality, but none seem to bounded above by anything.
Any sort of help at all would be greatly appreciated.