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Question Number 106410 by bemath last updated on 05/Aug/20

lim_(x→0) ((cos 2x.(1+sin^2 x)+cos^2 x−2)/x^2 )

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

Answered by john santu last updated on 05/Aug/20

lim_(x→0) ((cos 2x(2−cos^2 x)+cos^2 x−2)/x^2 )= lim_(x→0) (((2cos^2 x−1)(2−cos^2 x)−(2−cos^2 x))/x^2 )= lim_(x→0) (((2−cos^2 x){2cos^2 x−1−1})/x^2 )= lim_(x→0) ((2(cos^2 x−1)(2−cos^2 x))/x^2 )= lim_(x→0) ((2(−sin^2 x))/x^2 ) ×lim_(x→0) (2−cos^2 x)= −2 × 1 = −2

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

Answered by Dwaipayan Shikari last updated on 05/Aug/20

lim_(x→0) ((cos2x(1+sin^2 x)−(1+sin^2 x))/x^2 ) lim_(x→0) (((cos2x−1)(1+sin^2 x))/x^2 )=−2((sin^2 x)/x^2 ).(1+sin^2 x)=−2

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

Answered by mathmax by abdo last updated on 05/Aug/20

let use hospital thelrem u(x) =cos(2x)(1+sin^2 x)+cos^2 x−2 and v(x) =x^2 u(x) =cos(2x)(2−cos^2 x)+cos^2 x−2 =cos(2x){2−((1+cos(2x))/2)}+((1+cos(2x))/2)−2 =cos(2x){((3−cos(2x))/2)}+(1/2)cos(2x)−(3/2) =(3/2)cos(2x)−(1/2)(((1+cos(4x))/2))+(1/2)cos(2x)−(3/2) =2cos(2x)−(1/4)cos(4x)−(1/4)−(3/2) ⇒u^′ (x)=−4sin(2x) +sin(4x) and u^((2)) (x) =−8cos(2x)+4cos(2x) ⇒lim_(x→0) u^((2)) (x)=−4 v^′ (x)=2x and v^((2)) (x) =2 ⇒lim_(x→0) v^((2)) (x) =2 ⇒ lim_(x→0) ((u(x))/(v(x))) =−2

$$\mathrm{let}\:\mathrm{use}\:\mathrm{hospital}\:\mathrm{thelrem}\:\:\mathrm{u}\left(\mathrm{x}\right)\:=\mathrm{cos}\left(\mathrm{2x}\right)\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{v}\left(\mathrm{x}\right)\:=\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{u}\left(\mathrm{x}\right)\:=\mathrm{cos}\left(\mathrm{2x}\right)\left(\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2}\: \\ $$$$=\mathrm{cos}\left(\mathrm{2x}\right)\left\{\mathrm{2}−\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\right\}+\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}−\mathrm{2} \\ $$$$=\mathrm{cos}\left(\mathrm{2x}\right)\left\{\frac{\mathrm{3}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\right\}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$=\mathrm{2cos}\left(\mathrm{2x}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\left(\mathrm{4x}\right)−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\mathrm{u}^{'} \left(\mathrm{x}\right)=−\mathrm{4sin}\left(\mathrm{2x}\right)\:+\mathrm{sin}\left(\mathrm{4x}\right) \\ $$$$\mathrm{and}\:\mathrm{u}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\:=−\mathrm{8cos}\left(\mathrm{2x}\right)+\mathrm{4cos}\left(\mathrm{2x}\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{u}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)=−\mathrm{4} \\ $$$$\mathrm{v}^{'} \left(\mathrm{x}\right)=\mathrm{2x}\:\mathrm{and}\:\mathrm{v}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\:=\mathrm{2}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{v}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\:=\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{u}\left(\mathrm{x}\right)}{\mathrm{v}\left(\mathrm{x}\right)}\:=−\mathrm{2} \\ $$

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