## Calculus: Early Transcendentals 8th Edition

(a) We could estimate that the value of $\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1} \approx 0.7$ (b) We could guess that $\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1} = \frac{2}{3}$ (c) $\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1} = \frac{2}{3}$
(a) On the graph of $\frac{x}{\sqrt{1+3x}-1}$, the y-intercept is approximately 0.7 We could estimate that the value of $\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1} \approx 0.7$ (b) We can evaluate the function as $x$ approaches 0: $f(-0.1) = \frac{-0.1}{\sqrt{1+3(-0.1)}-1} = 0.61222$ $f(-0.01) = \frac{-0.01}{\sqrt{1+3(-0.01)}-1} = 0.66163$ $f(-0.001) = \frac{-0.001}{\sqrt{1+3(-0.001)}-1} = 0.66617$ $f(-0.0001) = \frac{-0.0001}{\sqrt{1+3(-0.0001)}-1} = 0.66662$ $f(-0.00001) = \frac{-0.00001}{\sqrt{1+3(-0.00001)}-1} = 0.666662$ $f(0.1) = \frac{0.1}{\sqrt{1+3(0.1)}-1} = 0.7134$ $f(0.01) = \frac{0.01}{\sqrt{1+3(0.01)}-1} = 0.67163$ $f(0.001) = \frac{0.001}{\sqrt{1+3(0.001)}-1} = 0.66716$ $f(0.0001) = \frac{0.0001}{\sqrt{1+3(0.0001)}-1} = 0.6667$ $f(0.00001) = \frac{0.00001}{\sqrt{1+3(0.00001)}-1} = 0.66667$ We can see that the value of the function is getting closer to $\frac{2}{3}$ as $x$ approaches 0. We could guess that $\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1} = \frac{2}{3}$ (c) We can evaluate the limit using limit laws: $\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1}$ $=\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1}\cdot \frac{\sqrt{1+3x}+1}{\sqrt{1+3x}+1}$ $=\lim\limits_{x \to 0}\frac{x\sqrt{1+3x}+1}{1+3x-1}$ $=\lim\limits_{x \to 0}\frac{x\sqrt{1+3x}+1}{3x}$ $=\lim\limits_{x \to 0}\frac{\sqrt{1+3x}+1}{3}$ $=\frac{\sqrt{1+0}+1}{3}$ $=\frac{2}{3}$