Chapter 2 - Limits - Review Exercises - Page 125: 44

$$\eqalign{ & \left( {\text{a}} \right)\frac{9}{4},\frac{9}{4} \cr & \left( {\text{b}} \right){\text{There is no slant asymptote}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{9{x^2} + 4}}{{{{\left( {2x - 1} \right)}^2}}} \cr & {\text{Expand the denominator}} \cr & f\left( x \right) = \frac{{9{x^2} + 4}}{{4{x^2} - 4x + 1}} \cr & \left( {\text{a}} \right){\text{ Calculate }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr & \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{9{x^2} + 4}}{{4{x^2} - 4x + 1}} \cr & {\text{Divide the numerator and denominator by }}{x^2} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{9{x^2}}}{{{x^2}}} + \frac{4}{{{x^2}}}}}{{\frac{{4{x^2}}}{{{x^2}}} - \frac{{4x}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{9 + \frac{4}{{{x^2}}}}}{{4 - \frac{4}{x} + \frac{1}{{{x^2}}}}} \cr & = \frac{{\overbrace {\mathop {\lim }\limits_{x \to \infty } 9}^{{\text{approaches 9}}} + \overbrace {\mathop {\lim }\limits_{x \to \infty } \frac{4}{{{x^2}}}}^{{\text{approaches 0}}}}}{{\underbrace {\mathop {\lim }\limits_{x \to \infty } 4}_{{\text{approaches }}4} - \underbrace {\mathop {\lim }\limits_{x \to \infty } \frac{4}{x}}_{{\text{approaches 0}}} + \underbrace {\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}}}_{{\text{approaches 0}}}}} \cr & = \frac{{9 + 0}}{{4 - 0 + 0}} \cr & = \frac{9}{4} \cr & and \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{{9 + \frac{4}{{{x^2}}}}}{{4 - \frac{4}{x} + \frac{1}{{{x^2}}}}} \cr & = \frac{{\overbrace {\mathop {\lim }\limits_{x \to - \infty } 9}^{{\text{approaches 9}}} + \overbrace {\mathop {\lim }\limits_{x \to - \infty } \frac{4}{{{x^2}}}}^{{\text{approaches 0}}}}}{{\underbrace {\mathop {\lim }\limits_{x \to - \infty } 4}_{{\text{approaches }}4} - \underbrace {\mathop {\lim }\limits_{x \to - \infty } \frac{4}{x}}_{{\text{approaches 0}}} + \underbrace {\mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^2}}}}_{{\text{approaches 0}}}}} \cr & = \frac{{9 + 0}}{{4 - 0 + 0}} \cr & = \frac{9}{4} \cr & \cr & \left( {\text{b}} \right){\text{The slant asymptote occurs with rational functions only }} \cr & {\text{when the degree of the polynomial in the numerator exceeds }} \cr & {\text{the degree of the polynomial in the denominator by exactly 1.}} \cr & {\text{For this exercise, the degree of the numerator is equal to the}} \cr & {\text{degree of the denominator}}{\text{, then there is no slant asymptote}}{\text{.}} \cr} $$