Calculus: $\lim_{x \to 1} \frac{x^3 - 1}{x-1}$ Is the limit $0$ or $3$?
$x^3 -1$ can be $(x-1)(x^2 +x +1)$, with the transformation, the limit will be $3$?
Why cannot we just say $x$ is to be $1$, $x^3 -1$ is going to be $0$ so the limit is $0$?

Use $$\frac{x^3-1}{x-1}=x^2+x+1$$ for $x\neq1$.
Also, by the definition of the limit if $x$ is closed to $1$, so $x\neq1$.
