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Question Number 195618 by cortano12 last updated on 06/Aug/23

$$\:\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$

Answered by MM42 last updated on 06/Aug/23

0

$$\mathrm{0} \\ $$

Commented by cortano12 last updated on 06/Aug/23

wrong

$$\mathrm{wrong} \\ $$

Answered by ajfour last updated on 06/Aug/23

(0/0) form diff. N^r & D^r ((1−((cos x)/( (√(1+x))))+2(√(1+x))sin x)/(2x)) (((√(1+x))−cos x+2(1+x)sin x)/(2x(√(1+x)))) x=2t (((√(1+2t))−1+2sin^2 t+4(1+2t)sin tcos t)/(4t(√(1+2t)))) (((√(1+2t))−1)/( 4t(√(1+2t))))+(((sin t)/t))(((sin t+2(1+2t)cos t)/(2(√(1+2t))))) =(1/(2((√(1+2t))+1)(√(1+2t))))+1×(2/2) =(1/4)+1=(5/4)

$$\frac{\mathrm{0}}{\mathrm{0}}\:\:{form}\:\:\:{diff}.\:{N}^{{r}} \:\&\:{D}^{{r}} \\ $$$$\frac{\mathrm{1}−\frac{\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{1}+{x}}}+\mathrm{2}\sqrt{\mathrm{1}+{x}}\mathrm{sin}\:{x}}{\mathrm{2}{x}} \\ $$$$\frac{\sqrt{\mathrm{1}+{x}}−\mathrm{cos}\:{x}+\mathrm{2}\left(\mathrm{1}+{x}\right)\mathrm{sin}\:{x}}{\mathrm{2}{x}\sqrt{\mathrm{1}+{x}}} \\ $$$${x}=\mathrm{2}{t} \\ $$$$\frac{\sqrt{\mathrm{1}+\mathrm{2}{t}}−\mathrm{1}+\mathrm{2sin}\:^{\mathrm{2}} {t}+\mathrm{4}\left(\mathrm{1}+\mathrm{2}{t}\right)\mathrm{sin}\:{t}\mathrm{cos}\:{t}}{\mathrm{4}{t}\sqrt{\mathrm{1}+\mathrm{2}{t}}} \\ $$$$\frac{\sqrt{\mathrm{1}+\mathrm{2}{t}}−\mathrm{1}}{\:\mathrm{4}{t}\sqrt{\mathrm{1}+\mathrm{2}{t}}}+\left(\frac{\mathrm{sin}\:{t}}{{t}}\right)\left(\frac{\mathrm{sin}\:{t}+\mathrm{2}\left(\mathrm{1}+\mathrm{2}{t}\right)\mathrm{cos}\:{t}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{2}{t}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{1}+\mathrm{2}{t}}+\mathrm{1}\right)\sqrt{\mathrm{1}+\mathrm{2}{t}}}+\mathrm{1}×\frac{\mathrm{2}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$

Answered by horsebrand11 last updated on 06/Aug/23

lim_(x→0) ((x+2−2(√(1+x)) cos x)/x^2 ) = lim_(x→0) ((x^2 +4x+4−4(1+x)cos^2 x)/((x+2+2(√(1+x)) cos x )x^2 )) = lim_(x→0) ((x^2 +4x(1−cos^2 x)+4(1−cos^2 x))/(4x^2 )) = (1/4) + lim_(x→0) ((sin^2 x)/x) +lim_(x→0) ((4sin^2 x)/(4x^2 )) = (1/4) + 0 + 1= (5/4)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}\:\mathrm{cos}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{4}−\mathrm{4}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{x}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}\:\mathrm{cos}\:\mathrm{x}\:\right)\mathrm{x}^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)+\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{4x}^{\mathrm{2}} } \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{x}}\:+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{4x}^{\mathrm{2}} } \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\mathrm{0}\:+\:\mathrm{1}=\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$

Answered by mr W last updated on 06/Aug/23

=lim_(x→0) ((x+2−2(1+(x/2)−(x^2 /8)+...)(1−(x^2 /2)+...))/x^2 ) =lim_(x→0) ((x+2−(2+x−((5x^2 )/4)+o(x^2 )))/x^2 ) =lim_(x→0) ((x+2−2−x+((5x^2 )/4)+o(x^2 ))/x^2 ) =lim_(x→0) ((((5x^2 )/4)+o(x^2 ))/x^2 ) =(5/4)

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+\mathrm{2}−\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{8}}+...\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+...\right)}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+\mathrm{2}−\left(\mathrm{2}+{x}−\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}+{o}\left({x}^{\mathrm{2}} \right)\right)}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+\mathrm{2}−\mathrm{2}−{x}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}+{o}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}+{o}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}} \\ $$

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