divide numerator and denominator by the highest power of x in the denominator and proce\n lim_{x\rightarrow-i…

Explanation:

Step1: Identify highest power

The highest power of $x$ in the denominator $x^{2}+9x$ is $x^{2}$.

Step2: Divide numerator and denominator by $x^{2}$

[ \begin{align*} \lim_{x\rightarrow-\infty}\left(\frac{1 - x^{4}}{x^{2}+9x}\right)^{7}&=\lim_{x\rightarrow-\infty}\left(\frac{\frac{1}{x^{2}}-\frac{x^{4}}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{9x}{x^{2}}}\right)^{7}\ &=\lim_{x\rightarrow-\infty}\left(\frac{\frac{1}{x^{2}}-x^{2}}{1 + \frac{9}{x}}\right)^{7} \end{align*} ]

Step3: Find limits of individual terms

As $x\rightarrow-\infty$, $\lim_{x\rightarrow-\infty}\frac{1}{x^{2}} = 0$ and $\lim_{x\rightarrow-\infty}\frac{9}{x}=0$. So, $\lim_{x\rightarrow-\infty}\left(\frac{\frac{1}{x^{2}}-x^{2}}{1+\frac{9}{x}}\right)^{7}=\lim_{x\rightarrow-\infty}\left(\frac{0 - x^{2}}{1+0}\right)^{7}=\lim_{x\rightarrow-\infty}(-x^{2})^{7}$

Step4: Evaluate the limit

As $x\rightarrow-\infty$, $(-x^{2})^{7}=-\infty$ since $x^{2}\rightarrow+\infty$ as $x\rightarrow\pm\infty$ and raising to the 7 - th power (an odd power) keeps the negative sign.

Answer:

$-\infty$