Answer
(a) Estimate the value of $\lim\limits_{x \to 0} \frac{x}{\sqrt {1+3x }- 1}$ by graphing the function
b) Make a table of values of $f(x)$ for $x$ close to 0 and guess
the value of the limit.
(c) Use the Limit Laws to prove that your guess is correct.
Work Step by Step
$$\lim\limits_{x \to 0} \frac{x}{\sqrt {1+3x}-1}=\lim\limits_{x \to 0} \frac{x}{\sqrt {1+3x}-1} * \frac{\sqrt {1+3x}+1}{\sqrt {1+3x}+1}=
\lim\limits_{x \to 0} \frac{x(\sqrt {1+3x}+1)}{(1+3x)-1} = \lim\limits_{x \to 0} \frac{\sqrt {1+3x}+1}{3} =\frac{\sqrt {1+3(0)}+1}{3} = \frac{2}{3} $$
