Answer:
[tex]\large\boxed{\lim\limits_{x\to-\infty}\dfrac{1-x^4}{x^2+4x}=-\infty}[/tex]
Step-by-step explanation:
[tex]\lim\limits_{x\to-\infty}\dfrac{1-x^4}{x^2+4x}=\lim\limits_{x\to-\infty}\dfrac{\frac{1}{x^2}-\frac{x^4}{x^2}}{\frac{x^2}{x^2}+\frac{4x}{x^2}}=\lim\limits_{x\to-\infty}\dfrac{\frac{1}{x^2}-x^2}{1+\frac{4}{x}}=(*)\\\\\dfrac{1}{x^2}\xrightarrow{x\to-\infty}0\\\\-x^\xrightarrow{x\to-\infty}}-\infty\\\\1\xrightarrow{x\to-\infty}1\\\\\dfrac{4}{x}\xrightarrow{x\to-\infty}0\\\\(*)=\dfrac{0-\infty}{1+0}=-\infty[/tex]
