2021-10-15

Find the limit:
$\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{1-{x}^{3}}{{x}^{2}+7x}\right)}^{5}$

sovienesY

The given limit expression is $\underset{x\to -\mathrm{\infty }}{lim}\left({\left(\frac{1-{x}^{3}}{{x}^{2}+7x}\right)}^{5}\right)$
Find the limit as follows:
$\underset{x\to -\mathrm{\infty }}{lim}\left({\left(\frac{1-{x}^{3}}{{x}^{2}+7x}\right)}^{5}\right)={\left(\underset{x\to -\mathrm{\infty }}{lim}\left(\frac{1-{x}^{3}}{{x}^{2}+7x}\right)\right)}^{5}$
$={\left(\underset{x\to -\mathrm{\infty }}{lim}\left(\frac{\frac{1}{{x}^{2}}-\frac{{x}^{3}}{{x}^{2}}}{\frac{{x}^{2}}{{x}^{2}}+\frac{7x}{{x}^{2}}}\right)\right)}^{5}$
$=\left(\underset{x\to -\mathrm{\infty }}{lim}{\left(\frac{\frac{1}{{x}^{2}}-x}{1+\frac{7}{x}}\right)}^{5}$
$=\left(\frac{\underset{x\to -\mathrm{\infty }}{lim}\left(\frac{1}{{x}^{2}}-x\right)}{\underset{x\to -\mathrm{\infty }}{lim}\left(1+\frac{7}{x}\right)}\right)$
$={\left(\frac{\mathrm{\infty }}{1}\right)}^{5}$
$=\mathrm{\infty }$

Jeffrey Jordon