## Calculus (3rd Edition)

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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 $$\lim _{x \rightarrow -\infty} \frac{4x^3 }{x^4+1}=\lim _{x \rightarrow -\infty} \frac{12x^2 }{4x^3}=\lim _{x \rightarrow -\infty} \frac{3 }{x}=0.$$