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Question Number 172361 by Mikenice last updated on 25/Jun/22

Answered by mr W last updated on 26/Jun/22

=lim_(x→0) ((x−(x^3 /3)+o(x^5 )−(x+(x^3 /6)+o(x^5 )))/x^3 ) =lim_(x→0) ((−((1/3)+(1/6))x^3 +o(x^5 ))/x^3 ) =−((1/3)+(1/6)) =−(1/2)

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{o}\left({x}^{\mathrm{5}} \right)−\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{5}} \right)\right)}{{x}^{\mathrm{3}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}\right){x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{5}} \right)}{{x}^{\mathrm{3}} } \\ $$$$=−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by bagjagugum123 last updated on 26/Jun/22

What is o Sir?

$${What}\:{is}\:{o}\:{Sir}? \\ $$

Commented by mr W last updated on 26/Jun/22

o(x^5 ) means following terms are of 5^(th) order or higher.

$${o}\left({x}^{\mathrm{5}} \right)\:{means}\:{following}\:{terms}\:{are}\:{of} \\ $$$$\mathrm{5}^{{th}} \:{order}\:{or}\:{higher}. \\ $$

Commented by mr W last updated on 26/Jun/22

Commented by bagjagugum123 last updated on 26/Jun/22

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

Answered by mahdipoor last updated on 25/Jun/22

=lim_(x→0) (((1/(x^2 +1))−(1/( (√(1−x^2 )))))/(3x^2 )) =(((√(1−x^2 ))−(1+x^2 ))/(3x^2 (x^2 +1)(√(1−x^2 ))))= =(((1−x^2 )−(1+x^2 )^2 )/(3x^2 (x^2 +1)(√(1−x^2 ))((√(1−x^2 ))+(1+x^2 ))))= =((−x^2 (x^2 +3))/(3x^2 (x^2 +1)(√(1−x^2 ))((√(1−x^2 ))+(1+x^2 ))))= =−(1/2)

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}{\mathrm{3}{x}^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}= \\ $$$$=\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)}= \\ $$$$=\frac{−{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

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