Question Number 186617 by mustafazaheen last updated on 07/Feb/23
$$\underset{{x}\rightarrow\infty} {{lim}}\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt[{\mathrm{5}}]{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}}=?\:\:\:\:\:\:\:\:{with}\:{solution} \\ $$
Commented by aba last updated on 07/Feb/23
$$ \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{5}}]{\frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{5}} }{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{5}}]{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }=\infty \\ $$
Commented by aba last updated on 07/Feb/23
$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\:\sqrt[{\mathrm{5}}]{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{8}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{10}} }}}=\frac{\mathrm{1}}{\mathrm{0}}=\infty \\ $$
Answered by cortano1 last updated on 07/Feb/23
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{5}}]{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }\:=\:\infty \\ $$