## Calculus: Early Transcendentals 9th Edition

(a) When we use the zoom function on a graphing calculator, we could estimate that $\lim\limits_{x \to -\infty}(\sqrt{x^2+x+1}+x) = -0.5$ (b) We could guess that $\lim\limits_{x \to -\infty}(\sqrt{x^2+x+1}+x) = -0.5$ (c) $\lim\limits_{x \to -\infty}(\sqrt{x^2+x+1}+x) = -0.5$
(a) When we use the zoom function on a graphing calculator, we could estimate that $\lim\limits_{x \to -\infty}(\sqrt{x^2+x+1}+x) = -0.5$ (b) We can evaluate $f(x)$ for decreasing values of $x$: $f(-10) = \sqrt{(-10)^2+(-10)+1}+(-10) = -0.4606$ $f(-100) = \sqrt{(-100)^2+(-100)+1}+(-100) = -0.4962$ $f(-1000) = \sqrt{(-1000)^2+(-1000)+1}+(-1000) = -0.4996$ $f(-10,000) = \sqrt{(-10,000)^2+(-10,000)+1}+(-10,000) = -0.49996$ $f(-100,000) = \sqrt{(-100,000)^2+(-100,000)+1}+(-100,000) = -0.499996$ $f(-1,000,000) = \sqrt{(-1,000,000)^2+(-1,000,000)+1}+(-1,000,000) = -0.4999996$ We could guess that $\lim\limits_{x \to -\infty}(\sqrt{x^2+x+1}+x) = -0.5$ (c) We can calculate the value of the limit: $\lim\limits_{x \to -\infty}(\sqrt{x^2+x+1}+x)$ $= \lim\limits_{x \to -\infty}(\sqrt{x^2+x+1}+x)\cdot \frac{\sqrt{x^2+x+1}-x}{\sqrt{x^2+x+1}-x}$ $= \lim\limits_{x \to -\infty}\frac{(x^2+x+1)-x^2}{\sqrt{x^2+x+1}-x}$ $= \lim\limits_{x \to -\infty}\frac{x+1}{\sqrt{x^2+x+1}-x}$ $= \lim\limits_{x \to -\infty}\frac{(x+1)(\frac{1}{x})}{(\sqrt{x^2+x+1}-x)(\frac{1}{x})}$ $= \lim\limits_{x \to -\infty}\frac{1+\frac{1}{x}}{-\sqrt{x^2/x^2+x/x^2+1/x^2}-1}$ $= \lim\limits_{x \to -\infty}\frac{1+\frac{1}{x}}{-\sqrt{1+1/x+1/x^2}-1}$ $= \frac{1+0}{-\sqrt{1+0+0}-1}$ $= \frac{1}{-2}$ $= -0.5$