Answer
$\lim_{x\to1}\dfrac{x^{8}-1}{x^{5}-1}=\dfrac{8}{5}$
The graph is shown below:
Work Step by Step
$\lim_{x\to1}\dfrac{x^{8}-1}{x^{5}-1}$
Try to evaluate the limit applying direct substitution:
$\lim_{x\to1}\dfrac{x^{8}-1}{x^{5}-1}=\dfrac{1^{8}-1}{1^{5}-1}=\dfrac{0}{0}$ Indeterminate form
The limit could not be evaluated using direct substitution. Factor the numerator and the denominator of the function and simplify:
$\lim_{x\to1}\dfrac{x^{8}-1}{x^{5}-1}=\lim_{x\to1}\dfrac{(x^{4}-1)(x^{4}+1)}{(x-1)(x^{4}+x^{3}+x^{2}+x+1)}=...$
$...=\lim_{x\to1}\dfrac{(x^{4}+1)(x^{2}-1)(x^{2}+1)}{(x-1)(x^{4}+x^{3}+x^{2}+x+1)}=...$
$...=\lim_{x\to1}\dfrac{(x^{4}+1)(x^{2}+1)(x-1)(x+1)}{(x-1)(x^{4}+x^{3}+x^{2}+x+1)}=...$
$...=\lim_{x\to1}\dfrac{(x^{4}+1)(x^{2}+1)(x+1)}{x^{4}+x^{3}+x^{2}+x+1}=...$
Try to evaluate the limit using direct substitution again:
$\lim_{x\to1}\dfrac{(x^{4}+1)(x^{2}+1)(x+1)}{x^{4}+x^{3}+x^{2}+x+1}=...$
$...=\dfrac{[(1^{4}+1)][(1^{2}+1)][(1+1)]}{1^{4}+1^{3}+1^{2}+1+1}=\dfrac{(2)(2)(2)}{5}=\dfrac{8}{5}$
