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lim-x-0-1-xsin-x-cos-2x-tg-2-x-2-




Question Number 26294 by d.monhbayr@gmail.com last updated on 23/Dec/17
lim_(x→0) (((√(1+xsin x ))−(√(cos 2x)))/(tg^2 (x/2)))
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{x}\mathrm{sin}\:{x}\:}−\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{{tg}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$
Commented by abdo imad last updated on 24/Dec/17
for x∈v(0)  (√(1+u))∼1+((u )/2)  tanu∼u  cosu∼1−(u^2 /2)   ⇒ (√(1+xsinx))∼1+((xsinx)/2)  ∼1+(x^2 /2)  tan^2 ((x/2))∼ (x^2 /4)  cos(2x)∼1−2x^(2 )  and (√(cos2x))∼1−x^2   ⇒(((√(1+xsinx))−(√(cos2x)))/(tan^2 ((x/2))))  ∼((1+(x^2 /2)−1+x^2 )/(x^2 /4))  =(((3/2)x^2 )/(x^2 /4))⇒lim_(x→0) (((3/2)x^2 )/(x^2 /4))=6
$${for}\:{x}\in{v}\left(\mathrm{0}\right)\:\:\sqrt{\mathrm{1}+{u}}\sim\mathrm{1}+\frac{{u}\:}{\mathrm{2}} \\ $$$${tanu}\sim{u} \\ $$$${cosu}\sim\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\Rightarrow\:\sqrt{\mathrm{1}+{xsinx}}\sim\mathrm{1}+\frac{{xsinx}}{\mathrm{2}}\:\:\sim\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\sim\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${cos}\left(\mathrm{2}{x}\right)\sim\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}\:} \:{and}\:\sqrt{{cos}\mathrm{2}{x}}\sim\mathrm{1}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{1}+{xsinx}}−\sqrt{{cos}\mathrm{2}{x}}}{{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\:\:\sim\frac{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}+{x}^{\mathrm{2}} }{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$=\frac{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} }{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} }{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}=\mathrm{6} \\ $$

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