Answer
$$\frac{9}{7}$$
Work Step by Step
We have
$$
\lim _{x \rightarrow 1} \frac{(1+3 x)^{1 / 2}-2}{(1+7 x)^{1 / 3}-2}=\frac{2-2}{2-2}=\frac{0}{0}
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 1} \frac{\frac{3}{2}(1+3 x)^{-1 / 2}}{\frac{7}{3}(1+7 x)^{-2 / 3}}
=\frac{3/4}{7/12}=\frac{9}{7}.
$$
