Answer
$$53$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^9} + 3{x^8} + 4{x^5} - 8}}{{x - 1}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}1{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 1} \left( {{x^9} + 3{x^8} + 4{x^5} - 8} \right) = {\left( 1 \right)^9} + 3{\left( 1 \right)^8} + 4{\left( 1 \right)^5} - 8 = 0 \cr
& \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{{D_x}\left( {{x^9} + 3{x^8} + 4{x^5} - 8} \right)}}{{{D_x}\left( {x - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{9{x^8} + 24{x^7} + 20{x^4}}}{1} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = 9{\left( 1 \right)^8} + 24{\left( 1 \right)^7} + 20{\left( 1 \right)^4} \cr
& = 9 + 24 + 20 \cr
& = 53 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^9} + 3{x^8} + 4{x^5} - 8}}{{x - 1}} = 53 \cr} $$
