Answer
$$\frac{4}{ 7}$$
Work Step by Step
Given $$\lim _{x \rightarrow - \infty}\frac{x+4x^3}{1-x^2+7x^3 }$$
Then
\begin{align*}
\lim _{x \rightarrow - \infty}\frac{x+4x^3}{1-x^2+7x^3 }&=\lim _{x \rightarrow - \infty}\frac{x/x^3+4x^3/x^3}{1/x^3-x^2/x^3+7x^3/x^3 }\\
&=\lim _{x \rightarrow - \infty}\frac{1/x^2+4 }{1/x^3-1/x +7 }\\
&= \frac{0+4}{0-0+7}\\
&= \frac{4}{ 7}
\end{align*}
