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