Answer
$\frac{7}{3}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^7} - 1}}{{{x^3} - 1}} \cr
& {\text{Use the quotient property of limits}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^7} - 1}}{{{x^3} - 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} \left( {{x^7} - 1} \right)}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^3} - 1} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to 1} \left( {{x^7}} \right) - \mathop {\lim }\limits_{x \to 1} \left( 1 \right)}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^3}} \right) - \mathop {\lim }\limits_{x \to 1} \left( 1 \right)}} \cr
& {\text{Evaluate the limit}} \cr
& \frac{{\mathop {\lim }\limits_{x \to 1} \left( {{x^7}} \right) - \mathop {\lim }\limits_{x \to 1} \left( 1 \right)}}{{\mathop {\lim }\limits_{x \to 1} \left( {{x^3}} \right) - \mathop {\lim }\limits_{x \to 1} \left( 1 \right)}} = \frac{{{{\left( 1 \right)}^7} - 1}}{{{{\left( 1 \right)}^3} - 1}}{\text{ = }}\frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^7} - 1}}{{{x^3} - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left[ {{x^7} - 1} \right]}}{{\frac{d}{{dx}}\left[ {{x^3} - 1} \right]}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{7{x^6}}}{{3{x^2}}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{7{x^4}}}{3} \cr
& {\text{Evaluate the limit when }}x \to 1 \cr
& = \frac{{7{{\left( 1 \right)}^4}}}{3} \cr
& = \frac{7}{3} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^7} - 1}}{{{x^3} - 1}} = \frac{7}{3} \cr} $$
