Question Number 105661 by bobhans last updated on 30/Jul/20
$$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\mathrm{tan}\:^{\mathrm{2}} {x}\:\sqrt{\left({x}−\frac{\pi}{\mathrm{2}}\right)−\mathrm{cos}\:\:^{\mathrm{2}} {x}}\:? \\ $$
Commented by Dwaipayan Shikari last updated on 30/Jul/20
$${limit}\:{doesn}'{t}\:{exist} \\ $$
Answered by bemath last updated on 30/Jul/20
$${set}\:{x}\:=\:\frac{\pi}{\mathrm{2}}+{m} \\ $$$$\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{m}−\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}+{m}\right)}}{\mathrm{cot}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{2}}+{m}\right)} \\ $$$$\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{m}−\mathrm{cos}\:^{\mathrm{2}} \left({m}\right)}}{\mathrm{tan}\:^{\mathrm{2}} {m}}=\:\sqrt{{m}}\:\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}−\frac{\mathrm{cos}\:^{\mathrm{2}} {m}}{{m}}}}{\mathrm{tan}\:^{\mathrm{2}} {m}} \\ $$$$=\sqrt{{m}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\frac{\mathrm{cos}\:^{\mathrm{2}} {m}}{\mathrm{2}{m}}}{\mathrm{tan}\:^{\mathrm{2}} {m}}=\:\sqrt{{m}}\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\left(\mathrm{1}−\frac{{m}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }{} \\ $$
Answered by Dwaipayan Shikari last updated on 30/Jul/20
$$\underset{{x}\rightarrow\frac{\pi^{−} }{\mathrm{2}}} {\mathrm{lim}}{tan}^{\mathrm{2}} {x}\sqrt{{x}−\frac{\pi}{\mathrm{2}}−{h}−{cos}^{\mathrm{2}} {x}}\:\:\:\:\rightarrow{i}\infty\:\:\:\left\{\left({h}\rightarrow\mathrm{0}\right)\left({L}.{H}.{S}\right)\right. \\ $$$$\left({R}.{H}.{S}\right)\underset{{x}\rightarrow\frac{\pi^{+} }{\mathrm{2}}} {\mathrm{lim}}{tan}^{\mathrm{2}} {x}\sqrt{{x}−\frac{\pi}{\mathrm{2}}+{h}−{cos}^{\mathrm{2}} {x}}\rightarrow\infty \\ $$$${R}.{H}.{S}\neq{L}.{H}.{S} \\ $$