Answer
$$\frac{2}{3}$$
Work Step by Step
Given $$c_{n}=\frac{\ln \left(n^{2}+1\right)}{\ln \left(n^{3}+1\right)}$$
Then
\begin{align*}
\lim_{n\to \infty } c_n &= \lim_{n\to \infty } \frac{\ln \left(n^{2}+1\right)}{\ln \left(n^{3}+1\right)}\ \ \\ & \text{Using L'Hopital's rule } \\
&=\lim _{n\to \infty \:}\left(\frac{\frac{2n}{n^2+1}}{\frac{3n^2}{n^3+1}}\right)\\
&=\lim _{n\to \infty \:}\left(\frac{2\left(n^3+1\right)}{3n\left(n^2+1\right)}\right)\\
&=\lim _{n\to \infty \:}\left(\frac{2n^2}{3n^2+1}\right) \\
&= \lim _{n\to \infty \:}\left(\frac{4n}{6n}\right)\\
&=\frac{2}{3}
\end{align*}
