Calculus: Early Transcendentals 9th Edition

(a) $\lim\limits_{x \to 0}\frac{\sqrt{3+x}-\sqrt{3}}{x} \approx 0.30$ (b) $\lim\limits_{x \to 0}\frac{\sqrt{3-0.1}-\sqrt{3}}{-0.1} = 0.2887$ (c) $\lim\limits_{x \to 0} \frac{\sqrt{3+x}-\sqrt{3}}{x} = \frac{1}{2\sqrt{3}}$
(a) On the graph of $\frac{\sqrt{3+x}-\sqrt{3}}{x}$, the y-intercept is approximately 0.30 We could estimate that the value of $\lim\limits_{x \to 0}\frac{\sqrt{3+x}-\sqrt{3}}{x} \approx 0.30$ (b) We can evaluate the function as $x$ approaches 0: $f(-0.1) = \frac{\sqrt{3-0.1}-\sqrt{3}}{-0.1} = 0.2911$ $f(-0.01) = \frac{\sqrt{3-0.01}-\sqrt{3}}{-0.01} = 0.2887$ $f(-0.001) = \frac{\sqrt{3-0.001}-\sqrt{3}}{-0.001} = 0.2887$ $f(-0.0001) = \frac{\sqrt{3-0.0001}-\sqrt{3}}{-0.0001} = 0.2887$ $f(-0.00001) = \frac{\sqrt{3-0.00001}-\sqrt{3}}{-0.00001} = 0.2887$ $f(0.1) = \frac{\sqrt{3+0.1}-\sqrt{3}}{0.1} = 0.2863$ $f(0.01) = \frac{\sqrt{3+0.01}-\sqrt{3}}{0.01} = 0.2884$ $f(0.001) = \frac{\sqrt{3+0.001}-\sqrt{3}}{0.001} = 0.2887$ $f(0.0001) = \frac{\sqrt{3+0.0001}-\sqrt{3}}{0.0001} = 0.2887$ $f(0.00001) = \frac{\sqrt{3+0.00001}-\sqrt{3}}{0.00001} = 0.2887$ We can see that the value of the function is getting closer to $0.2887$ as $x$ approaches 0. We could estimate that $\lim\limits_{x \to 0}\frac{\sqrt{3-0.1}-\sqrt{3}}{-0.1} = 0.2887$ (c) We can evaluate the limit using limit laws: $\lim\limits_{x \to 0} \frac{\sqrt{3+x}-\sqrt{3}}{x}$ $=\lim\limits_{x \to 0} \frac{\sqrt{3+x}-\sqrt{3}}{x} \cdot \frac{\sqrt{3+x}+\sqrt{3}}{\sqrt{3+x}+\sqrt{3}}$ $=\lim\limits_{x \to 0}\frac{3+x-3}{x(\sqrt{3+x}+\sqrt{3})}$ $=\lim\limits_{x \to 0}\frac{x}{x(\sqrt{3+x}+\sqrt{3})}$ $=\lim\limits_{x \to 0}\frac{1}{(\sqrt{3+x}+\sqrt{3})}$ $=\frac{1}{2\sqrt{3}}$