Given, $\lim _{x \rightarrow 1} \frac{x^3-1}{x-1}$
If we put $x=1$, then expression $\frac{x^3-1}{x-1}$ becomes the indeterminate form $\frac{0}{0}$.
Therefore, $(x-1)$ is a common factor of $\left(x^3-1\right)$ and $(x-1)$.
Factorising the numerator and denominator, we have
$\lim _{x \rightarrow 1} \frac{x^3-1}{x-1}\left[\frac{0}{0}\right.$ form $]$
$=\lim _{x \rightarrow 1} \frac{(x-1)\left(x^2+x+1\right)}{(x-1)}$
$=\lim _{x \rightarrow 1}\left(x^2+x+1\right)$
$=1^2+1+1$
$=3$
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