Evaluate

To evaluate:

$\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}$

Formula used: We have,

$\lim _{x \rightarrow a} f(x)=f(a)$ and

$x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$

As $\mathrm{X} \rightarrow 1$, we have

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

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

$\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}=1+1+1$

$\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}=3$

Thus, the value of $\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}$ is 3 .