Question Number 39078 by Cheyboy last updated on 02/Jul/18
$$\boldsymbol{{Without}}\:\boldsymbol{{using}}\:\boldsymbol{{l}}'\boldsymbol{{hopital}} \\ $$$$\boldsymbol{{find}}\:\:\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}−\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{x}}−\mathrm{3}} \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 03/Jul/18
$${lim}_{{x}\rightarrow\mathrm{3}^{−} } \:\:\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}−\mathrm{3}}\:\:={lim}_{{x}\rightarrow\mathrm{3}^{−} } \:\:−\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{\mathrm{3}−{x}}\:\:\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{3}^{−} } \:\:\:−\sqrt{\frac{\mathrm{9}−{x}^{\mathrm{2}} }{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }}={lim}_{{x}\rightarrow\mathrm{3}^{−} } \:\:−\sqrt{\frac{\mathrm{3}+{x}}{\mathrm{3}−{x}}} \\ $$$$=\:−\infty\:. \\ $$
Commented by orlandorap123 last updated on 03/Aug/18
$$ \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}−\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{x}}−\mathrm{3}}\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\left(\mathrm{3}−{x}\right)\left(\mathrm{3}+{x}\right)}{−\left({x}−\mathrm{3}\right)\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\left(\mathrm{3}+{x}\right)}{−\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }} \\ $$$$\mathrm{evaluando} \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\left(\mathrm{6}\right)}{−\mathrm{0}}=−\infty \\ $$$$\boldsymbol{{find}}\:\:\:\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{9}−\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{x}}−\mathrm{3}}=−\infty \\ $$
Answered by ajfour last updated on 02/Jul/18
$${left}\:{hand}\:{limit}=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{9}−\left(\mathrm{3}−{h}\right)^{\mathrm{2}} }}{\left(\mathrm{3}−{h}\right)−\mathrm{3}}\: \\ $$$$\:\:\:\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{6}{h}−{h}^{\mathrm{2}} }}{−{h}}\:=\:−\infty \\ $$
Commented by Cheyboy last updated on 02/Jul/18
$${Thank}\:{you}\:{sir} \\ $$