divide numerator and denominator by the highest power of x in the denominator\n lim_{x\rightarrow-infty}left(…

divide numerator and denominator by the highest power of x in the denominator\n lim_{x\rightarrow-infty}left(\frac{1 - x^{4}}{x^{2}+6x}\right)^{7} \n lim_{x\rightarrow-infty}left(\frac{1 - x^{4}}{x^{2}+6x}\right)^{7}=square \text{ (simplify your answer.)}
Answer
Explanation:
Step1: Identify highest power
The highest power of $x$ in the denominator $x^{2}+6x$ is $x^{2}$.
Step2: Divide numerator and denominator by $x^{2}$
[ \begin{align*} \lim_{x\rightarrow-\infty}\left(\frac{1 - x^{4}}{x^{2}+6x}\right)^{7}&=\lim_{x\rightarrow-\infty}\left(\frac{\frac{1}{x^{2}}-\frac{x^{4}}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{6x}{x^{2}}}\right)^{7}\ &=\lim_{x\rightarrow-\infty}\left(\frac{\frac{1}{x^{2}}-x^{2}}{1 + \frac{6}{x}}\right)^{7} \end{align*} ]
Step3: Evaluate limits of individual terms
As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{1}{x^{2}} = 0$ and $\lim_{x\rightarrow-\infty}\frac{6}{x}=0$. So we have $\lim_{x\rightarrow-\infty}\left(\frac{0 - x^{2}}{1+0}\right)^{7}=\lim_{x\rightarrow-\infty}(-x^{2})^{7}$.
Step4: Determine the result
As $x\rightarrow-\infty$, $(-x^{2})^{7}=-\infty$.
Answer:
$-\infty$