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