Answer
$${\text{The limit does not exist}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{8{x^6} + 3{x^4} - 9x}}{{9{x^7} - 2{x^4} + {x^3}}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {8{x^6} + 3{x^4} - 9x} \right) = 8{\left( 0 \right)^6} + 3{\left( 0 \right)^4} - 9\left( 0 \right) = 0 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {9{x^7} - 2{x^4} + {x^3}} \right) = 9{\left( 0 \right)^7} - 2{\left( 0 \right)^4} + {\left( 0 \right)^3} = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{D_x}\left( {8{x^6} + 3{x^4} - 9x} \right)}}{{{D_x}\left( {9{x^7} - 2{x^4} + {x^3}} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{48{x^5} + 12{x^3} - 9}}{{63{x^6} - 8{x^3} + 3{x^2}}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{48{{\left( 0 \right)}^5} + 12{{\left( 0 \right)}^3} - 9}}{{63{{\left( 0 \right)}^6} - 8{{\left( 0 \right)}^3} + 3{{\left( 0 \right)}^2}}} \cr
& = \frac{{ - 9}}{0} \cr
& {\text{Since the limits does not leads to the indeterminate form }}\frac{0}{0}{\text{ or }}\frac{{ \pm \infty }}{{ \pm \infty }} \cr
& {\text{We cannot apply l'Hospital's rule}}. \cr
& {\text{}} \cr
& {\text{The limit does not exist}} \cr} $$
