The graph of $f(x)$ near vertical asymptote

Question

Suppose $f(x) = \frac{x}{x^3+x+1} $ . I want to understand graph of it near vertical asymptote . It's obvious that we should solve $x^3+x+1 = 0 $ but it is too hard . Also I know there is a way for doing it without solving that . The derivative of $x^3+x+1$ is $3x^2 + 1$ and this expression is always positive . Therefore we can conclude $\lim_{x \to a^{+}} x^3 + x + 1 = 0^{+} $ and $\lim_{x \to a^{-}} x^3 + x + 1 = 0^{-} $ where $a$ is the root of $x^3 + x + 1$ . Also $a$ is a negative number because $x^3+x+1$ is strickly increasing function and when $x=0$ then value of it is $1$ . From these conclusions we can easily grasp the graph but I'm looking for the other methods or ideas that can help me to solve questions like this .

Answer 1

The equation $x^3+x+1=0$ has one real solution $\alpha\doteq-0.682328$. Do the polynomial division $$(x^3+x+1):(x-\alpha)=x^2+\alpha x+\alpha^2+1\ ,$$ treating $\alpha$ as name of the real number satisfying $\alpha^3+\alpha+1=0$. You then can separate $f$ into partial fractions $$f(x):={x\over x^3+x+1}={A\over x-\alpha}+{Bx+C\over x^2+\alpha x+\alpha^2+1}\ .\tag{1}$$ Determine $A$, $B$, $C$ in the usual way as functions of $\alpha$. The second term on the RHS of $(1)$ is a nice function near $x=\alpha$, whereas the term ${A\over x-\alpha}$ governs the asymptotic behavior of $f$ near $x=\alpha$. Note that apart from determining $\alpha$ once and for all no numerical work has to be done here.

Answer 2

The vertical asymptote is the real root of the denominator:

$x=a=(\frac{\sqrt{93}}{18}-\frac{1}{2})^{1/3}- (\frac{\sqrt{93}}{18}+\frac{1}{2})^{1/3}$.

Now to know the trend around $x=a$, you have to calculate the limit of the function $f(x)$ for $x$ tends to $a$ from the right and left, that is:

$\lim_{x \to a^{+}} f(x) = -\infty$,

$\lim_{x \to a^{-}} f(x) = +\infty$.