Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises - Page 97: 55

$$\eqalign{ & a.\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 3;\,\,\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = 3;\,\,\,\,\,\,y = 3 \cr & b.\,x = 4;\,\,\,\,\,\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right) = - \infty ;\,\,\,\,\,\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right) = + \infty \cr & \,\,\,\,\,x = - 3;\,\,\,\,\,\mathop {\lim }\limits_{x \to - {3^ - }} f\left( x \right) = + \infty ;\,\,\,\,\,\mathop {\lim }\limits_{x \to - {3^ + }} f\left( x \right) = - \infty \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{3{x^4} + 3{x^3} - 36{x^2}}}{{{x^4} - 25{x^2} + 144}} \cr & {\text{Factor the numerator and denominator}} \cr & f\left( x \right) = \frac{{3{x^2}\left( {{x^2} + x - 12} \right)}}{{\left( {{x^2} - 16} \right)\left( {{x^2} - 9} \right)}} \cr & f\left( x \right) = \frac{{3{x^2}\left( {x + 4} \right)\left( {x - 3} \right)}}{{\left( {x + 4} \right)\left( {x - 4} \right)\left( {x + 3} \right)\left( {x - 3} \right)}} \cr & {\text{Cancel the common factors}} \cr & f\left( x \right) = \frac{{3{x^2}}}{{\left( {x - 4} \right)\left( {x + 3} \right)}} \cr & f\left( x \right) = \frac{{3{x^2}}}{{{x^2} - x - 12}}, x\ne-4,3 \cr & \cr & f\left( x \right) = \frac{{\frac{{3{x^2}}}{{{x^2}}}}}{{\frac{{{x^2}}}{{{x^2}}} - \frac{x}{{{x^2}}} - \frac{{12}}{{{x^2}}}}} \cr & f\left( x \right) = \frac{3}{{1 - \frac{1}{x} - \frac{{12}}{{{x^2}}}}} \cr & \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 } = \frac{3}{{1 - \frac{1}{\infty } - \frac{{12}}{{{\infty ^2}}}}} = 3 \cr & \mathop {\lim }\limits_{x \to \infty } = \frac{3}{{1 - \frac{1}{{ - \infty }} - \frac{{12}}{{{{\left( { - \infty } \right)}^2}}}}} = 3 \cr & {\text{,then the horizontal asymptote of }}f\left( x \right){\text{ is }}y = 3 \cr & \cr & b.\,\,{\text{The function is not defined for }}\left( {x - 4} \right)\left( {x + 3} \right) = 0 \cr & \,\,\,\,\,{\text{there are a vertical asymptotes at }}x = 4{\text{ and }}x = - 3 \cr & \cr & {\text{Analyze }}\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right) \cr & \mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {4^ - }} \frac{{3{x^2}}}{{{x^2} + x - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3{{\left( 4 \right)}^2}}}{{{{\left( 4 \right)}^2} + \left( 4 \right) - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{48}}{{{0^ - }}} = - \infty \cr & \mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {4^ + }} \frac{{3{x^2}}}{{{x^2} + x - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3{{\left( 4 \right)}^2}}}{{{{\left( 4 \right)}^2} + \left( 4 \right) - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{48}}{{{0^ - }}} = \infty \cr & \cr & {\text{Analyze }}\mathop {\lim }\limits_{x \to - {3^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - {3^ + }} f\left( x \right) \cr & \mathop {\lim }\limits_{x \to - {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {3^ - }} \frac{{3{x^2}}}{{{x^2} + x - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3{{\left( { - 3} \right)}^2}}}{{{{\left( { - 3} \right)}^2} + \left( { - 3} \right) - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{48}}{{{0^ + }}} = \infty \cr & \mathop {\lim }\limits_{x \to - {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {3^ + }} \frac{{3{x^2}}}{{{x^2} + x - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3{{\left( { - 3} \right)}^2}}}{{{{\left( { - 3} \right)}^2} + \left( { - 3} \right) - 12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{48}}{{{0^ - }}} = - \infty \cr} $$