Answer
$\lim_{x\to1}\dfrac{x^{3}-1}{x^{2}-1}=\dfrac{3}{2}$
Work Step by Step
$\lim_{x\to1}\dfrac{x^{3}-1}{x^{2}-1}$
Try to evaluate the limit applying direct substitution:
$\lim_{x\to1}\dfrac{x^{3}-1}{x^{2}-1}=\dfrac{1^{3}-1}{1^{2}-1}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Factor the numerator and the denominator of the function and simplify:
$\lim_{x\to1}\dfrac{x^{3}-1}{x^{2}-1}=\lim_{x\to1}\dfrac{(x-1)(x^{2}+x+1)}{(x-1)(x+1)}=...$
$...=\lim_{x\to1}\dfrac{x^{2}+x+1}{x+1}=...$
Try direct substitution again:
$...=\lim_{x\to1}\dfrac{x^{2}+x+1}{x+1}=\dfrac{1^{2}+1+1}{1+1}=\dfrac{3}{2}$
