Question Number 144564 by imjagoll last updated on 26/Jun/21
Commented by justtry last updated on 26/Jun/21
$$\mathrm{0}?? \\ $$
Answered by liberty last updated on 26/Jun/21
$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{2021}} }{\mathrm{x}^{\mathrm{2021}} }\right)+\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{5}}\right)^{\mathrm{2021}} }{\mathrm{x}^{\mathrm{2021}} }\right) \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{1}−\mathrm{2x}^{−\mathrm{1}} +\mathrm{x}^{−\mathrm{2}} }}{\mathrm{x}}\right)^{\mathrm{2021}} +\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{1}+\mathrm{5x}^{−\mathrm{2}} }}{\mathrm{x}}\right)^{\mathrm{2021}} \\ $$$$=\mathrm{2}^{\mathrm{2021}} +\mathrm{2}^{\mathrm{2021}} \:=\:\mathrm{2}^{\mathrm{2022}} \\ $$
Commented by justtry last updated on 26/Jun/21
$${nice},{thank}\:{you} \\ $$