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 $0$?
You have shown that you can see why the limit of the expression should be $3$. Regarding why it shouldn't be $0$, consider how $\displaystyle \frac{x^3-1}{x-1}=\left(x^2+x+1\right)\frac{x-1}{x-1}$ would be evaluated at different $x$-values. At $x=3$, we have $\require{cancel}\displaystyle \left(3^2+3+1\right)\frac{\cancel{2}}{\cancel{2}}$ and likewise, at $x=0.5$, we would have $\require{cancel}\displaystyle \left(0.5^2+0.5+1\right)\frac{\cancel{0.5}}{\cancel{0.5}}$. The pattern should be clear; for any $x\ne1$, the terms in the fraction are nonzero and can cancel with no problem. But this includes all $x$ in the vicinity of $1$, which is exactly what we pass through as we approach $1$. Since a limit is not interested in what happens at $x=1$ but in what happens near $x=1$, and we never need to hit $1$ exactly, we never need to evaluate $\cfrac00$ and it does not affect the answer.
Since x can not be 1，and x^3-1=(x-1)(x^2+x+1) ,we can transform the former one to x^2 + x + 1. The limit of it is 1+1+1=3.
For why it can’t be 0. The limit of (x-1) is also 0. We can’t make a conclusion that how much is the limit of 0/0.