Chapter 2 - Limits - Review Exercises - Page 124: 42

$$\eqalign{ & {\text{Vertical asymptotes}}:{\text{ }}x = - 2,{\text{ }}x = \frac{1}{2} \cr & {\text{Horizontal asymptote: }}y = 1 \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{2{x^2} + 6}}{{2{x^2} + 3x - 2}} \cr & {\text{*To calculate the vertical asymptotes we need to find the values }} \cr & {\text{at which the function is undefined}}{\text{. Set the denominator to 0}} \cr & 2{x^2} + 3x - 2 = 0 \cr & \left( {x + 2} \right)\left( {2x - 1} \right) = 0 \cr & x = - 2,{\text{ }}x = \frac{1}{2} \cr & {\text{Then}}{\text{, the vertical asymptotes are: }}x = - 2,{\text{ }}x = \frac{1}{2} \cr & \cr & {\text{*Evaluate }}\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{{2{x^2} + 6}}{{2{x^2} + 3x - 2}} \cr & {\text{Divide the numerator and denominator by }}{x^2} \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{2{x^2}}}{{{x^2}}} + \frac{6}{{{x^2}}}}}{{\frac{{2{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} - \frac{2}{{{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to - \infty } \frac{{2 + \frac{6}{{{x^2}}}}}{{2 + \frac{3}{x} - \frac{2}{{{x^2}}}}} \cr & = \frac{{\overbrace {\mathop {\lim }\limits_{x \to - \infty } \left( {2 + \frac{6}{{{x^2}}}} \right)}^{{\text{approaches 2}}}}}{{\underbrace {\mathop {\lim }\limits_{x \to - \infty } \left( {2 + \frac{3}{x} - \frac{2}{{{x^2}}}} \right)}_{{\text{approaches 2}}}}} = \frac{2}{2} = 1 \cr & and \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{2{x^2}}}{{{x^2}}} + \frac{6}{{{x^2}}}}}{{\frac{{2{x^2}}}{{{x^2}}} + \frac{{3x}}{{{x^2}}} - \frac{2}{{{x^2}}}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{2 + \frac{6}{{{x^2}}}}}{{2 + \frac{3}{x} - \frac{2}{{{x^2}}}}} \cr & = \frac{{\overbrace {\mathop {\lim }\limits_{x \to \infty } \left( {2 + \frac{6}{{{x^2}}}} \right)}^{{\text{approaches 2}}}}}{{\underbrace {\mathop {\lim }\limits_{x \to \infty } \left( {2 + \frac{3}{x} - \frac{2}{{{x^2}}}} \right)}_{{\text{approaches 2}}}}} = \frac{2}{2} = 1 \cr & {\text{Then}}{\text{, }}y = 1{\text{ is a horizontal asymptote}}{\text{.}} \cr & \cr & {\text{Summary:}} \cr & {\text{Vertical asymptotes}}:{\text{ }}x = - 2,{\text{ }}x = \frac{1}{2} \cr & {\text{Horizontal asymptote: }}y = 1 \cr} $$