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 $]$ $\begin{array}{l}=\lim _{x \rightarrow 1} \frac{(x-1)\left(x^2+x+1\right)}{(x-1)} \\ =\lim _{x \rightarrow 1}\left(x^2+x+1\right)=1^2+1+1=3\end{array}$
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