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Question Number 39078 by Cheyboy last updated on 02/Jul/18

Without using l′hopital find lim_(x→3) ((√(9−x^2 ))/(x−3))

$$\boldsymbol{{Without}}\:\boldsymbol{{using}}\:\boldsymbol{{l}}'\boldsymbol{{hopital}} \\ $$$$\boldsymbol{{find}}\:\:\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}−\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{x}}−\mathrm{3}} \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 03/Jul/18

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

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

Commented by orlandorap123 last updated on 03/Aug/18

lim_(x→3) ((√(9−x^2 ))/(x−3)) lim_(x→3) (((3−x)(3+x))/(−(x−3)(√(9−x^2 )))) lim_(x→3) (((3+x))/(−(√(9−x^2 )))) evaluando lim_(x→3) (((6))/(−0))=−∞ find lim_(x→3) ((√(9−x^2 ))/(x−3))=−∞

$$ \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}−\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{x}}−\mathrm{3}}\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\left(\mathrm{3}−{x}\right)\left(\mathrm{3}+{x}\right)}{−\left({x}−\mathrm{3}\right)\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\left(\mathrm{3}+{x}\right)}{−\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$$$\mathrm{evaluando} \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\left(\mathrm{6}\right)}{−\mathrm{0}}=−\infty \\ $$$$\boldsymbol{{find}}\:\:\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}−\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{x}}−\mathrm{3}}=−\infty \\ $$

Answered by ajfour last updated on 02/Jul/18

left hand limit= lim_(h→0) ((√(9−(3−h)^2 ))/((3−h)−3)) =lim_(h→0) ((√(6h−h^2 ))/(−h)) = −∞

$${left}\:{hand}\:{limit}=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{9}−\left(\mathrm{3}−{h}\right)^{\mathrm{2}} }}{\left(\mathrm{3}−{h}\right)−\mathrm{3}}\: \\ $$$$\:\:\:\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{6}{h}−{h}^{\mathrm{2}} }}{−{h}}\:=\:−\infty \\ $$

Commented by Cheyboy last updated on 02/Jul/18

Thank you sir

$${Thank}\:{you}\:{sir} \\ $$

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