## Calculus 8th Edition

Published by Cengage

# Chapter 3 - Applications of Differentiation - 3.4 Limits at Infinity; Horizontal Asymptotes - 3.4 Exercises - Page 242: 13

#### Answer

$$1$$

#### Work Step by Step

Given $$\lim _{t\rightarrow \infty}\frac{\sqrt{t}+t^2}{2t-t^2}$$ Then \begin{aligned} \lim _{t\rightarrow \infty}\frac{\sqrt{t}+t^2}{2t-t^2} &= \lim _{t\rightarrow \infty}\frac{\frac{\sqrt{t}}{t^2}+\frac{t^2}{t^2}}{\frac{2t}{t^2}-\frac{t^2}{t^2}} \\ &= \lim _{t\rightarrow \infty}\frac{\frac{1}{t^{3/2}}+1}{\frac{2 }{t }-1} \\ &= \lim _{x\rightarrow \infty}\frac{0+1 }{0-1}\\ &=1 \end{aligned}

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