Answer
$\lim\limits_{x \to +\infty} \frac{(2x-1)^5}{(3x^2+2x-7)(x^3-9x)} = \frac{32}{3}$
Work Step by Step
$\lim\limits_{x \to +\infty} \frac{(2x-1)^5}{(3x^2+2x-7)(x^3-9x)}$
Note that the degree on the numerator and denominator are both five.
Because we are taking the limit to positive infinity and the degree in the numerator and denominator are equal, we only need to consider the coefficients of the leading terms in the numerator and denominator. Thus:
$\lim\limits_{x \to +\infty} \frac{32x^5}{3x^5} = \frac{32}{3}$
