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Question Number 138996 by bramlexs22 last updated on 21/Apr/21
lim_(x→π/2) (((√2)−(√(1+sin x)))/( (√2) cos^2 x)) =?
$$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}}{\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=? \\ $$$$ \\ $$
Answered by EDWIN88 last updated on 21/Apr/21
lim_(x→π/2) ((1−sin x)/( (√2)(1−sin x))).lim_(x→π/2) (1/((1+sin x)((√2)+(√(1+sin x))))) = (1/( (√2))) . (1/(2(2(√2)))) = (1/8)
$$\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)}.\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by mathmax by abdo last updated on 22/Apr/21
let f(x)=(((√2)−(√(1+sinx)))/( (√2)cos^2 x)) ⇒f(x)=_(x=(π/2)−t) (((√2)−(√(1+cost)))/( (√2)sin^2 t))=g(t) (x→(π/2) ⇒t→0) cost ∼1−(t^2 /2) ⇒1+cost∼2−(t^2 /2) and sint∼t (√(1+cost))∼(√(2−(t^2 /2)))=(√2)(√(1−(t^2 /4)))∼(√2)(1−(t^2 /8)) ⇒ g(t)∼(((√2)−(√2)(1−(t^2 /8)))/( (√2)t^2 ))=(1/8) ⇒lim_(t→0) g(t)=(1/8)=lim_(x→(π/2)) f(x)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}+\mathrm{sinx}}}{\:\sqrt{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=_{\mathrm{x}=\frac{\pi}{\mathrm{2}}−\mathrm{t}} \:\:\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}+\mathrm{cost}}}{\:\sqrt{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \mathrm{t}}=\mathrm{g}\left(\mathrm{t}\right) \\ $$$$\left(\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\right)\:\:\mathrm{cost}\:\sim\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{1}+\mathrm{cost}\sim\mathrm{2}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{sint}\sim\mathrm{t} \\ $$$$\sqrt{\mathrm{1}+\mathrm{cost}}\sim\sqrt{\mathrm{2}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}}=\sqrt{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{4}}}\sim\sqrt{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{8}}\right)\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{t}\right)\sim\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{8}}\right)}{\:\sqrt{\mathrm{2}}\mathrm{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \mathrm{f}\left(\mathrm{x}\right) \\ $$

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