Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises - Page 97: 54

$$\eqalign{ & a.\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \frac{5}{2};\,\,\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \frac{5}{2};\,\,\,\,\,\,y = \frac{5}{2} \cr & b.\,x = - \sqrt 2 ;\,\,\,\,\,\mathop {\lim }\limits_{x \to - {{\sqrt 2 }^ - }} f\left( x \right) = \infty ;\,\,\,\,\,\mathop {\lim }\limits_{x \to - {{\sqrt 2 }^ + }} f\left( x \right) = - \infty \cr & \,\,\,\,\,x = \sqrt 2 ;\,\,\,\,\,\mathop {\lim }\limits_{x \to {{\sqrt 2 }^ - }} f\left( x \right) = - \infty ;\,\,\,\,\,\mathop {\lim }\limits_{x \to {{\sqrt 2 }^ + }} f\left( x \right) = \infty \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{\sqrt {16{x^4} + 64{x^2}} + {x^2}}}{{2{x^2} - 4}} \cr & a.{\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{{\sqrt {16{x^4} + 64{x^2}} + {x^2}}}{{2{x^2} - 4}} \cr & {\text{Using properties of limits}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\overbrace {\sqrt {16\mathop {\lim }\limits_{x \to \infty } {x^4} + 64\mathop {\lim }\limits_{x \to \infty } {x^2}} }^{{\text{approaches }}\infty } + \overbrace {\mathop {\lim }\limits_{x \to \infty } {x^2}}^{{\text{approaches }}\infty }}}{{\underbrace {2\mathop {\lim }\limits_{x \to \infty } {x^2} - \mathop {\lim }\limits_{x \to \infty } \left( 4 \right)}_{{\text{approaches }}\infty }}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{\infty }{\infty } \cr & {\text{Divide de numerator and denominator by }}{x^2} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {16{x^4} + 64{x^2}} + {x^2}}}{{2{x^2} - 4}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {\frac{{16{x^4}}}{{{x^4}}} + \frac{{64{x^2}}}{{{x^4}}}} + \frac{{{x^2}}}{{{x^2}}}}}{{\frac{{2{x^2}}}{{{x^2}}} - \frac{4}{{{x^2}}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {16 + \frac{{64}}{{{x^2}}}} + 1}}{{2 - \frac{4}{{{x^2}}}}} \cr & {\text{Using properties of limits}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {16 + \frac{{64}}{{{x^2}}}} + 1}}{{2 - \frac{4}{{{x^2}}}}} = \frac{{\sqrt {\mathop {\lim }\limits_{x \to \infty } \left( {16} \right) + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{64}}{{{x^2}}}} \right)} + \mathop {\lim }\limits_{x \to \infty } \left( 1 \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( 2 \right) - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{4}{{{x^2}}}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt {16 + 0} + 1}}{{2 - 0}} = \frac{5}{2} \cr & and \cr & \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \frac{{\sqrt {\mathop {\lim }\limits_{x \to - \infty } \left( {16} \right) + \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{64}}{{{x^2}}}} \right)} + 1}}{{\mathop {\lim }\limits_{x \to - \infty } \left( 2 \right) - \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{4}{{{x^2}}}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt {16 + 0} + 1}}{{2 - 0}} = \frac{5}{2} \cr & {\text{Then, the horizontal asymptote of }}f\left( x \right){\text{ is }}y = \frac{5}{2} \cr & \cr & b.\,\,{\text{The function is not defined for }}2{x^2} - 4 = 0 \cr & \,\,\,\,\,2{x^2} = 4 \cr & \,\,\,\,\,\,\,\,{x^2} = 2 \cr & \,\,\,\,\,\,\,\,\,\,x = \pm \sqrt 2 \cr & \,\,\,\,\,{\text{Then, there is a vertical asymptote at }}x = - \sqrt 2 {\text{ and }}x = \sqrt 2 \cr & \cr & {\text{Analyze }}\mathop {\lim }\limits_{x \to - {{\sqrt 2 }^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - {{\sqrt 2 }^ + }} f\left( x \right) \cr & \mathop {\lim }\limits_{x \to - {{\sqrt 2 }^ - }} \left( {\frac{{\sqrt {16 + \frac{{64}}{{{{\left( { - \sqrt 2 } \right)}^2}}}} + 1}}{{2 - \frac{4}{{{{\left( { - \sqrt 2 } \right)}^2}}}}}} \right) = + \infty \cr & \mathop {\lim }\limits_{x \to - {{\sqrt 2 }^ + }} \left( {\frac{{\sqrt {16 + \frac{{64}}{{{{\left( { - \sqrt 2 } \right)}^2}}}} + 1}}{{2 - \frac{4}{{{{\left( { - \sqrt 2 } \right)}^2}}}}}} \right) = - \infty \cr & \cr & {\text{Analyze }}\mathop {\lim }\limits_{x \to {{\sqrt 2 }^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to {{\sqrt 2 }^ + }} f\left( x \right) \cr & \mathop {\lim }\limits_{x \to {{\sqrt 2 }^ - }} \left( {\frac{{\sqrt {16 + \frac{{64}}{{{{\left( {\sqrt 2 } \right)}^2}}}} + 1}}{{2 - \frac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}}}} \right) = - \infty \cr & \mathop {\lim }\limits_{x \to {{\sqrt 2 }^ + }} \left( {\frac{{\sqrt {16 + \frac{{64}}{{{{\left( {\sqrt 2 } \right)}^2}}}} + 1}}{{2 - \frac{4}{{{{\left( {\sqrt 2 } \right)}^2}}}}}} \right) = \infty \cr} $$