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Question Number 111326 by bemath last updated on 03/Sep/20
    (√(bemath))    lim_(x→0)  ((1/(x sin^(−1) (x))) − (1/x^2 ) ) =?
$$\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{x}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)}\:−\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:\right)\:=? \\ $$
Answered by john santu last updated on 03/Sep/20
      JS^(−−−−−−−−) ★  let sin^(−1) (x)= p ⇒x = sin p    lim_(p→0)  ((1/(p sin p)) − (1/(sin^2 p))) =     lim_(p→0) (((sin p−p)/(psin^2 p))) =   lim_(p→0)  (((cos p−1)/(3p^2 )))= lim_(p→0)  (((−2sin^2 ((p/2)))/(3p^2 )))   = −(2/3)×(1/4) = −(1/6)
$$\:\:\:\:\:\:{J}\overset{−−−−−−−−} {{S}}\bigstar \\ $$$${let}\:\mathrm{sin}^{−\mathrm{1}} \left({x}\right)=\:{p}\:\Rightarrow{x}\:=\:\mathrm{sin}\:{p} \\ $$$$\:\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{p}\:\mathrm{sin}\:{p}}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} {p}}\right)\:=\: \\ $$$$\:\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{p}−{p}}{{p}\mathrm{sin}\:^{\mathrm{2}} {p}}\right)\:= \\ $$$$\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{cos}\:{p}−\mathrm{1}}{\mathrm{3}{p}^{\mathrm{2}} }\right)=\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{p}}{\mathrm{2}}\right)}{\mathrm{3}{p}^{\mathrm{2}} }\right)\: \\ $$$$=\:−\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{4}}\:=\:−\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by bemath last updated on 03/Sep/20
Answered by Dwaipayan Shikari last updated on 03/Sep/20
lim_(x→0) ((1/(x(x+(x^3 /6))))−(1/(x.x)))=(1/x)(((x−x−(x^3 /6))/(x(x+(x^3 /6)))))=((−(x/6))/(x+(x^3 /6)))=−(1/6)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)}−\frac{\mathrm{1}}{{x}.{x}}\right)=\frac{\mathrm{1}}{{x}}\left(\frac{{x}−{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)}\right)=\frac{−\frac{{x}}{\mathrm{6}}}{{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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