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Question Number 170141 by sciencestudent last updated on 17/May/22

lim_(x→0) ((1/(sin^2 x))−(1/x^2 ))=?

$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=? \\ $$

Commented by mr W last updated on 17/May/22

=((1/(sin x))+(1/x))((1/(sin x))−(1/x)) =((x/(sin x))+1)(((x−sin x)/x^3 ))(x/(sin x)) =((x/(sin x))+1)(((x−x+(x^3 /(3!))−(x^5 /(5!))+...)/x^3 ))(x/(sin x)) =((x/(sin x))+1)((1/(3!))−(x^2 /(5!))+...)(x/(sin x)) =(1+1)×((1/(3!)))×1 =(2/(3!)) =(1/3)

$$=\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{1}}{{x}}\right)\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\left(\frac{{x}}{\mathrm{sin}\:{x}}+\mathrm{1}\right)\left(\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\right)\frac{{x}}{\mathrm{sin}\:{x}} \\ $$$$=\left(\frac{{x}}{\mathrm{sin}\:{x}}+\mathrm{1}\right)\left(\frac{{x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+...}{{x}^{\mathrm{3}} }\right)\frac{{x}}{\mathrm{sin}\:{x}} \\ $$$$=\left(\frac{{x}}{\mathrm{sin}\:{x}}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{{x}^{\mathrm{2}} }{\mathrm{5}!}+...\right)\frac{{x}}{\mathrm{sin}\:{x}} \\ $$$$=\left(\mathrm{1}+\mathrm{1}\right)×\left(\frac{\mathrm{1}}{\mathrm{3}!}\right)×\mathrm{1} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by sciencestudent last updated on 17/May/22

How did You simplipy ((1/(sinx))−(1/x))((1/(sinx))+(1/x))?

$${How}\:{did}\:{You}\:{simplipy}\:\left(\frac{\mathrm{1}}{{sinx}}−\frac{\mathrm{1}}{{x}}\right)\left(\frac{\mathrm{1}}{{sinx}}+\frac{\mathrm{1}}{{x}}\right)? \\ $$

Commented by mr W last updated on 17/May/22

a^2 −b^2 =(a−b)(a+b)

$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({a}−{b}\right)\left({a}+{b}\right) \\ $$

Answered by ajfour last updated on 17/May/22

l=lim_(x→0) ((2/(1−cos 2x))−(1/x^2 )) =lim_(x→0) ((2/(1−[1−((4x^2 )/(2!))+((16x^4 )/(4!))−..]))−(1/x^2 )) =lim_(x→0) (1/x^2 )((/(1−((1x^2 )/3)+sx^4 −..))−1) =lim_(x→0) ((((1/3)−sx^2 +...)/(1−(x^2 /3)+sx^4 −...)))=(1/3)

$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{1}−\left[\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{16}{x}^{\mathrm{4}} }{\mathrm{4}!}−..\right]}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{}{\mathrm{1}−\frac{\mathrm{1}{x}^{\mathrm{2}} }{\mathrm{3}}+{sx}^{\mathrm{4}} −..}−\mathrm{1}\right) \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\frac{\mathrm{1}}{\mathrm{3}}−{sx}^{\mathrm{2}} +...}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+{sx}^{\mathrm{4}} −...}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by sciencestudent last updated on 17/May/22

How to get 1−(x^2 /3)+sx^4 −∙∙∙?

$${How}\:{to}\:{get}\:\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+{sx}^{\mathrm{4}} −\centerdot\centerdot\centerdot? \\ $$

Commented by mr W last updated on 17/May/22

sin (x)=x−(x^3 /(3!))+(x^5 /(5!))−... cos (x)=1−(x^2 /(2!))+(x^4 /(4!))−... read more about Maclaurin series...

$$\mathrm{sin}\:\left({x}\right)={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−... \\ $$$$\mathrm{cos}\:\left({x}\right)=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}−... \\ $$$${read}\:{more}\:{about}\:{Maclaurin}\:{series}... \\ $$

Commented by mr W last updated on 17/May/22

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