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Question Number 194891 by cortano12 last updated on 18/Jul/23

lim_(x→3^+ ) ((((√x)−(√(x−3))−(√3))/( (√(x^2 −9)))) )=?

$$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{3}^{+} } {\mathrm{lim}}\:\left(\frac{\sqrt{{x}}−\sqrt{{x}−\mathrm{3}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}\:\right)=? \\ $$

Answered by MM42 last updated on 18/Jul/23

lim_(x→3) ((((√x)−(√3))/( (√(x−3))×(√(x+3)))) − ((√(x−3))/( (√(x−3))×(√(x+3)))) ) =lim_(x→3) ((((√(x−3))×(√(x−3)))/( ((√x)+(√3))×(√(x−3))×(√(x+3)))) − ((√(x−3))/( (√(x−3))×(√(x+3)))) ) =lim_(x→3) (((√(x−3))/( ((√x)+(√3))×(√(x+3)))) − (1/( (√(x+3)))) ) =−(1/( (√3))) ✓

$${lim}_{{x}\rightarrow\mathrm{3}} \:\left(\frac{\sqrt{{x}}−\sqrt{\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:−\:\frac{\sqrt{{x}−\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{3}} \:\left(\frac{\sqrt{{x}−\mathrm{3}}×\sqrt{{x}−\mathrm{3}}}{\:\left(\sqrt{{x}}+\sqrt{\mathrm{3}}\right)×\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:−\:\frac{\sqrt{{x}−\mathrm{3}}}{\:\sqrt{{x}−\mathrm{3}}×\sqrt{{x}+\mathrm{3}}}\:\right) \\ $$$$={lim}_{{x}\rightarrow\mathrm{3}} \:\left(\frac{\sqrt{{x}−\mathrm{3}}}{\:\left(\sqrt{{x}}+\sqrt{\mathrm{3}}\right)×\sqrt{{x}+\mathrm{3}}}\:−\:\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{3}}}\:\right) \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\checkmark \\ $$$$ \\ $$

Commented by cortano12 last updated on 18/Jul/23

= −(1/( (√3) )) (×) = −(1/( (√6))) (✓)

$$\:\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:}\:\left(×\right) \\ $$$$\:\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\:\left(\checkmark\right) \\ $$

Commented by MM42 last updated on 18/Jul/23

ok it was my mistake x→3⇒ −(1/( (√(x+3))))=−(1/( (√6)))

$${ok} \\ $$$${it}\:{was}\:{my}\:{mistake} \\ $$$${x}\rightarrow\mathrm{3}\Rightarrow\:−\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{3}}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}} \\ $$

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