Chapter 2 - Limits - Review Exercises - Page 125: 43

$$\eqalign{ & \left( {\text{a}} \right)\infty , - \infty \cr & \left( {\text{b}} \right)\frac{3}{4}x + \frac{5}{{16}}{\text{ is the slant asymptote}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{3{x^2} + 2x - 1}}{{4x + 1}} \cr & \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{{3{x^2} + 2x - 1}}{{4x + 1}} \cr & {\text{Divide the numerator and denominator by }}{x^2} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{3{x^2}}}{{{x^2}}} + \frac{{2x}}{{{x^2}}} - \frac{1}{{{x^2}}}}}{{\frac{{4x}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{3 + \frac{2}{x} - \frac{1}{{{x^2}}}}}{{\frac{4}{x} + \frac{1}{{{x^2}}}}} \cr & = \frac{{\overbrace {\mathop {\lim }\limits_{x \to \infty } 3}^{{\text{approaches }}3} + \overbrace {\mathop {\lim }\limits_{x \to \infty } \frac{2}{x}}^{{\text{approaches 0}}} - \overbrace {\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}}}^{{\text{approaches 0}}}}}{{\underbrace {\mathop {\lim }\limits_{x \to \infty } \frac{4}{x}}_{{\text{approaches }}{{\text{0}}^ + }} + \underbrace {\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}}}_{{\text{approaches }}{{\text{0}}^ + }}}} \cr & = \frac{{3 + 0 - 0}}{{0 + 0}} = \frac{3}{{{0^ + }}} \cr & = + \infty \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} + 2x - 1}}{{4x + 1}} = + \infty \cr & and \cr & \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{3{x^2} + 2x - 1}}{{4x + 1}} \cr & {\text{Divide the numerator and denominator by }}{x^2} \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{3{x^2}}}{{{x^2}}} + \frac{{2x}}{{{x^2}}} - \frac{1}{{{x^2}}}}}{{\frac{{4x}}{{{x^2}}} + \frac{1}{{{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{{3 + \frac{2}{x} - \frac{1}{{{x^2}}}}}{{\frac{4}{x} + \frac{1}{{{x^2}}}}} \cr & = \frac{{\overbrace {\mathop {\lim }\limits_{x \to - \infty } 3}^{{\text{approaches }}3} + \overbrace {\mathop {\lim }\limits_{x \to - \infty } \frac{2}{x}}^{{\text{approaches 0}}} - \overbrace {\mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^2}}}}^{{\text{approaches 0}}}}}{{\underbrace {\mathop {\lim }\limits_{x \to - \infty } \frac{4}{x}}_{{\text{approaches }}{{\text{0}}^ - }} + \underbrace {\mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^2}}}}_{{\text{approaches }}{{\text{0}}^ + }}}} \cr & = \frac{{3 + 0 - 0}}{{0 + 0}} = \frac{3}{{{0^ - }}} \cr & = - \infty \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} + 2x - 1}}{{4x + 1}} = - \infty \cr & \cr & \left( {\text{b}} \right)f\left( x \right) = \frac{{3{x^2} + 2x - 1}}{{4x + 1}} \cr & {\text{Using the long division}}{\text{, the function }}f\left( x \right){\text{ can be written as}} \cr & f\left( x \right) = \frac{{3{x^2} + 2x - 1}}{{4x + 1}} = \overbrace {\frac{3}{4}x + \frac{5}{{16}}}^{\ell \left( x \right)} - \frac{{21/16}}{{4x + 1}} \cr & {\text{As }}x{\text{ tends to infinite}}{\text{, the term }}\frac{{21/16}}{{4x + 1}}{\text{ approaches to 0}} \cr & {\text{and }}\mathop {\lim }\limits_{x \to \infty } \ell \left( x \right) = \infty ,{\text{ }}\mathop {\lim }\limits_{x \to - \infty } \ell \left( x \right) = - \infty ,{\text{ then}} \cr & {\text{The line }}\ell \left( x \right) = \frac{3}{4}x + \frac{5}{{16}}{\text{ is a slant asymptote}}{\text{.}} \cr} $$