divide numerator and denominator by the highest power of x in the denominator and proceed from there. find…

divide numerator and denominator by the highest power of x in the denominator and proceed from there. find the following limit. write ∞ or - ∞ where appropriate.\n lim_{x\rightarrow-infty}left(\frac{1 - x^{4}}{x^{2}+9x}\right)^{5}\n lim_{x\rightarrow-infty}left(\frac{1 - x^{4}}{x^{2}+9x}\right)^{5}=square \text{ (simplify your answer.)}

divide numerator and denominator by the highest power of x in the denominator and proceed from there. find the following limit. write ∞ or - ∞ where appropriate.\n lim_{x\rightarrow-infty}left(\frac{1 - x^{4}}{x^{2}+9x}\right)^{5}\n lim_{x\rightarrow-infty}left(\frac{1 - x^{4}}{x^{2}+9x}\right)^{5}=square \text{ (simplify your answer.)}

Answer

Explanation:

Step1: Identify highest power

The highest power of $x$ in the denominator $x^{2}+9x$ is $x^{2}$. Divide numerator and denominator of the fraction inside the parentheses by $x^{2}$. $\lim_{x\rightarrow-\infty}\left(\frac{\frac{1}{x^{2}}-\frac{x^{4}}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{9x}{x^{2}}}\right)^{5}=\lim_{x\rightarrow-\infty}\left(\frac{\frac{1}{x^{2}} - x^{2}}{1+\frac{9}{x}}\right)^{5}$

Step2: Evaluate 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 we have $\left(\frac{0-\lim_{x\rightarrow-\infty}x^{2}}{1 + 0}\right)^{5}=\left(-\lim_{x\rightarrow-\infty}x^{2}\right)^{5}$

Step3: Find the overall limit

Since $\lim_{x\rightarrow-\infty}x^{2}=\infty$, then $(-\lim_{x\rightarrow-\infty}x^{2})^{5}=(-\infty)^{5}=-\infty$

Answer:

$-\infty$