Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises - Page 97: 56

$$\eqalign{ & a.\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = - 2;\,\,\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = - 2;\,\,\,\,\,\,y = - 2 \cr & b.\,{\text{ No vertical asymptotes}}{\text{.}} \cr} $$

$$\eqalign{ & f\left( x \right) = 16{x^2}\left( {4{x^2} - \sqrt {16{x^4} + 1} } \right) \cr & {\text{Rationalizing the function}} \cr & f\left( x \right) = 16{x^2}\left( {4{x^2} - \sqrt {16{x^4} + 1} } \right) \times \frac{{4{x^2} + \sqrt {16{x^4} + 1} }}{{4{x^2} + \sqrt {16{x^4} + 1} }} \cr & f\left( x \right) = \frac{{16{x^2}\left( {{{\left( {4{x^2}} \right)}^2} - {{\left( {\sqrt {16{x^4} + 1} } \right)}^2}} \right)}}{{4{x^2} + \sqrt {16{x^4} + 1} }} \cr & f\left( x \right) = \frac{{16{x^2}\left( {16{x^4} - 16{x^4} - 1} \right)}}{{4{x^2} + \sqrt {16{x^4} + 1} }} \cr & f\left( x \right) = - \frac{{16{x^2}}}{{4{x^2} + \sqrt {16{x^4} + 1} }} \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 } f\left( x \right) \cr & \mathop {\lim }\limits_{x \to \infty } \left( { - \frac{{16{x^2}}}{{4{x^2} + \sqrt {16{x^4} + 1} }}} \right) = - \frac{{16{{\left( \infty \right)}^2}}}{{4{{\left( \infty \right)}^2} + \sqrt {16{{\left( \infty \right)}^4} + 1} }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{\infty }{\infty } \cr & {\text{Divide the numerator and denominator by }}{x^2} \cr & \mathop {\lim }\limits_{x \to \infty } \left( { - \frac{{16{x^2}}}{{4{x^2} + \sqrt {16{x^4} + 1} }}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( { - \frac{{\frac{{16{x^2}}}{{{x^2}}}}}{{\frac{{4{x^2}}}{{{x^2}}} + \sqrt {\frac{{16{x^4}}}{{{x^4}}} + \frac{1}{{{x^4}}}} }}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to \infty } \left( { - \frac{{16}}{{4 + \sqrt {16 + \frac{1}{{{x^4}}}} }}} \right) \cr & {\text{calculate the limit}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( { - \frac{{16}}{{4 + \sqrt {16 + \frac{1}{{{x^4}}}} }}} \right) = - \frac{{16}}{{4 + \sqrt {16 + \frac{1}{{{{\left( \infty \right)}^4}}}} }} = - \frac{{16}}{{4 + 4}} = - 2 \cr & *\,\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr & \mathop {\lim }\limits_{x \to - \infty } \left( { - \frac{{16}}{{4 + \sqrt {16 + \frac{1}{{{x^4}}}} }}} \right) = - \frac{{16}}{{4 + \sqrt {16 + \frac{1}{{{{\left( { - \infty } \right)}^4}}}} }} = - \frac{{16}}{{4 + 4}} = - 2 \cr & {\text{,then the horizontal asymptote of }}f\left( x \right){\text{ is }}y = - 2 \cr & \cr & b.\,\,{\text{There is no vertical asymptote, because the denominator is}} \cr & {\text{always positive}}{\text{.}} \cr} $$