To my understanding, function $f(x)=\frac{1+x}{1+x^3}$ should not be defined with $x=-1$ since:
$f(-1)=\frac{1+(-1)}{1+(-1)^3}=\frac{1+(-1)}{1+(-1)}=\frac{0}{0}$
However, Google's plot drawer and Wolfram suggests a plot like this:
https://www.wolframalpha.com/input/?i=plot+%281%2Bx%29%2F%281%2Bx%5E3%29
With function $g(x)=\frac{1+2x}{1+x^3}$ it gives an expected result:
https://www.wolframalpha.com/input/?i=plot+%281%2B2x%29%2F%281%2Bx%5E3%29
Could someone explain what is happening? Am I missing something?
For $f(x)$ the function has a removable singularity at $x=-1$, as is apparent from the factorisation $$x^3+1=(x+1)(x^2-x+1)$$ This gives rise to the expression, valid for $x\neq -1$ $$f(x)=\frac 1{x^2-x+1}$$ If we then define $f(-1)$ using this alternative expression, we can plug the singularity and recover a continuous function. Without plugging the singularity we can see that near $x=-1$, $f(x)$ is a continuous function with a single point removed.
The same is not true for $g(x)$ (which does not, in any event, reduce to $\frac 00$ at $x=-1$)