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Mathematics 13 Online
OpenStudy (miaaziz):

how to evaluate lim x approach 0 ..(x-tan^-1)/x^3??

OpenStudy (jango_in_dtown):

@Miaaziz (x- arc tan (x))/x^3?

OpenStudy (miaaziz):

yes

OpenStudy (jango_in_dtown):

ok. here comes the solution

OpenStudy (jango_in_dtown):

OpenStudy (jango_in_dtown):

@Miaaziz

OpenStudy (miaaziz):

thank you . but how do you get (x-x^3/3 + x^5/5.....)/x^3 ?

OpenStudy (jango_in_dtown):

Thats the Gregory series expansion of arc tan x

OpenStudy (jango_in_dtown):

arc tan x= x-x^3/3 +x^5/5-x^7/7+....

OpenStudy (miaaziz):

oh i see. thanks . you help me a lot

OpenStudy (sshayer):

\[\lim_{x \rightarrow 0}\frac{ x-arc \tan x }{ x^3 },\frac{ 0 }{ 0 }~form\] \[=\lim_{x \rightarrow 0}\frac{ 1-\frac{ 1 }{ 1+x^2 } }{ 3x^2 }\] \[=\lim_{x \rightarrow 0}\frac{ 1+x^2-1 }{ \left( 1+x^2 \right)3x^2 }\] \[=\lim_{x \rightarrow 0}\frac{ x^2 }{ \left( 1+x^2 \right) 3x^2}\] \[=\lim_{x \rightarrow 0}\frac{ 1 }{ 3\left( 1+x^2 \right) }=\frac{ 1 }{ 3 }\]

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