Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 138: 21

$ - 1$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{4 - \sqrt x }}{{2 + \sqrt x }} \cr & {\text{Use properties of limits}} \cr & = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {4 - \sqrt x } \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {2 + \sqrt x } \right)}} \cr & = \frac{{4 - \sqrt \infty }}{{2 + \sqrt \infty }} = \frac{\infty }{\infty } \cr & {\text{Divide the numerator and denominator by }}\sqrt x \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{4 - \sqrt x }}{{2 + \sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{4}{{\sqrt x }} - \frac{{\sqrt x }}{{\sqrt x }}}}{{\frac{2}{{\sqrt x }} + \frac{{\sqrt x }}{{\sqrt x }}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{4}{{\sqrt x }} - 1}}{{\frac{2}{{\sqrt x }} + 1}} \cr & {\text{Use properties of limits}} \cr & = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{4}{{\sqrt x }}} \right) - \mathop {\lim }\limits_{x \to \infty } \left( 1 \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{2}{{\sqrt x }}} \right) + \mathop {\lim }\limits_{x \to \infty } \left( 1 \right)}} \cr & {\text{Evaluate the limits}} \cr & {\text{ = }}\frac{{0 - 1}}{{0 + 1}} \cr & = - 1 \cr & {\text{Therefore}}{\text{,}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{4 - \sqrt x }}{{2 + \sqrt x }} = - 1 \cr} $$