Question Number 109861 by john santu last updated on 26/Aug/20
$$\:\:\frac{{JS}}{\blacksquare\bigstar\bigstar\blacksquare} \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{5}−\mathrm{2}{x}+\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{4}+\sqrt{{x}−\mathrm{2}}}\:? \\ $$
Answered by bemath last updated on 26/Aug/20
Commented by john santu last updated on 26/Aug/20
$${beautifull} \\ $$
Answered by Her_Majesty last updated on 26/Aug/20
$$\frac{\mathrm{5}−\mathrm{2}{x}+\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{4}+\sqrt{{x}−\mathrm{2}}}=\frac{\mathrm{6}−\mathrm{2}{x}+\mathrm{3}\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{6}}\:\Rightarrow\:{lim}=−\mathrm{1} \\ $$
Commented by bemath last updated on 26/Aug/20
$${what}\:{your}\:{method}\:{miss}\:? \\ $$
Commented by Her_Majesty last updated on 26/Aug/20
$$\frac{\mathrm{5}−\mathrm{2}{x}+\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{4}+\sqrt{{x}−\mathrm{2}}}×\frac{{x}−\mathrm{4}−\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{4}−\sqrt{{x}−\mathrm{2}}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left(\mathrm{6}−\mathrm{2}{x}+\mathrm{3}\sqrt{{x}−\mathrm{2}}\right)}{\left({x}−\mathrm{3}\right)\left({x}−\mathrm{6}\right)}=\frac{\mathrm{6}−\mathrm{2}{x}+\mathrm{3}\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{6}} \\ $$
Commented by bemath last updated on 26/Aug/20
$${cooll}….{nice} \\ $$
Answered by 1549442205PVT last updated on 26/Aug/20
$$\underset{\mathrm{x}\rightarrow\mathrm{3}} {\mathrm{lim}\:}\frac{\mathrm{5}−\mathrm{2}{x}+\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{4}+\sqrt{{x}−\mathrm{2}}}\:\underset{\mathrm{L}'\mathrm{Hopital}} {=\:\:\:}\:\:\underset{\mathrm{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{2}}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{2}}}} \\ $$$$=\frac{−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}=\frac{−\mathrm{3}/\mathrm{2}}{\mathrm{3}/\mathrm{2}}=−\mathrm{1} \\ $$