Question Number 156683 by cortano last updated on 14/Oct/21
Commented by john_santu last updated on 14/Oct/21
$${it}\:{should}\:{be} \\ $$$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{1}+\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}−\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}{{x}−\mathrm{1}+\sqrt{{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}\:. \\ $$
Commented by MathSh last updated on 14/Oct/21
$$\mathrm{Square}\:\mathrm{root}\:\mathrm{stop}\:\mathrm{after}\:\mathrm{x}+\mathrm{1} \\ $$
Commented by cortano last updated on 15/Oct/21
$$\mathrm{yes}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}+\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}−\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}}{\mathrm{x}−\mathrm{1}+\sqrt{\mathrm{x}+\mathrm{1}}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\: \\ $$
Answered by john_santu last updated on 15/Oct/21
