Answer
(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) $\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1} = \frac{2}{3}$
Work Step by Step
(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:
$\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1}$
$=\lim\limits_{x \to 0}\frac{x}{\sqrt{1+3x}-1}\cdot \frac{\sqrt{1+3x}+1}{\sqrt{1+3x}+1}$
$=\lim\limits_{x \to 0}\frac{x\sqrt{1+3x}+1}{1+3x-1}$
$=\lim\limits_{x \to 0}\frac{x\sqrt{1+3x}+1}{3x}$
$=\lim\limits_{x \to 0}\frac{\sqrt{1+3x}+1}{3}$
$=\frac{\sqrt{1+0}+1}{3}$
$=\frac{2}{3}$
