Question Number 114754 by bemath last updated on 21/Sep/20
$${Without}\:{L}'{Hopital}\: \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:−\mathrm{2tan}\:{x}\:\right)}{\mathrm{cos}\:\mathrm{2}{x}}\:? \\ $$
Commented by bemath last updated on 21/Sep/20
$${gave}\:{kudos}\:{all}\:{sir} \\ $$
Answered by john santu last updated on 21/Sep/20
$${let}\:{x}\:=\:\frac{\pi}{\mathrm{4}}+{h}\:,\:{h}\rightarrow\mathrm{0} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+{h}\right)−\mathrm{2tan}\left(\frac{\pi}{\mathrm{4}}+{h}\right)\:}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{h}\right)}\:= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+{h}\right)−\mathrm{2tan}\:\left(\frac{\pi}{\mathrm{4}}+{h}\right)+\mathrm{1}}{−\mathrm{sin}\:\mathrm{2}{h}}\:= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+{h}\right)\right)^{\mathrm{2}} }{−\mathrm{sin}\:\mathrm{2}{h}}\:= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left[\mathrm{1}−\left(\frac{\mathrm{1}+\mathrm{tan}\:{h}}{\mathrm{1}−\mathrm{tan}\:{h}}\right)\right]^{\mathrm{2}} }{−\mathrm{sin}\:\mathrm{2}{h}}\:= \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(−\mathrm{2tan}\:{h}\right)^{\mathrm{2}} }{−\mathrm{sin}\:\mathrm{2}{h}}\:=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4}\:\mathrm{tan}\:^{\mathrm{2}} {h}}{−\mathrm{sin}\:\mathrm{2}{h}}\:=\:\mathrm{0} \\ $$
Answered by Olaf last updated on 21/Sep/20
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}−\mathrm{2tan}{x}}{\mathrm{cos2}{x}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\left(\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{\mathrm{cos2}{x}} \\ $$$${u}\:=\:\frac{\pi}{\mathrm{4}}−{x} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[\mathrm{1}−\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−{u}\right)^{\mathrm{2}} \right]^{\mathrm{2}} }{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{u}\right)} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[\mathrm{1}−\left(\frac{\mathrm{tan}\frac{\pi}{\mathrm{4}}−\mathrm{tan}{u}}{\mathrm{1}+\mathrm{tan}\frac{\pi}{\mathrm{4}}\mathrm{tan}{u}}\right)^{\mathrm{2}} \right]^{\mathrm{2}} }{\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{u}\right)} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{tan}{u}}{\mathrm{1}+\mathrm{tan}{u}}\right)^{\mathrm{2}} \right]^{\mathrm{2}} }{\mathrm{sin2}{u}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{tan}{u}}{\mathrm{1}+\mathrm{tan}{u}}\right)^{\mathrm{2}} \right]^{\mathrm{2}} }{\mathrm{sin2}{u}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[\frac{\mathrm{4tan}^{\mathrm{2}} {u}}{\left(\mathrm{1}+\mathrm{tan}{u}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} }{\mathrm{sin2}{u}} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[\frac{\mathrm{4}{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} }{\mathrm{2}{u}}\:=\:\mathrm{0} \\ $$
Answered by bobhans last updated on 21/Sep/20
$${in}\:{other}\:{way}\: \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{tan}\:{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}}\:=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)} \\ $$$$=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)}\:=\:\mathrm{0}\: \\ $$
Answered by MJS_new last updated on 21/Sep/20
$$\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x} \\ $$$$\mathrm{cos}\:\mathrm{2}{x}\:=−\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:{x}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)}{\mathrm{1}+\mathrm{tan}\:{x}}\:=\mathrm{0} \\ $$