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Question Number 42786 by maxmathsup by imad last updated on 02/Sep/18
calculate lim_(x→0)    ((1−(x/(sinx)))/x^2 )
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}−\frac{{x}}{{sinx}}}{{x}^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 30/Sep/18
we have sinx = x−(x^3 /(3!)) +o(x^3 )  (x→0) ⇒  ((sinx −x)/(x^2 sinx))  ∼  ((−(x^3 /(3!)))/x^3 ) →−(1/6)  when x→0 ⇒  lim_(x→0)  ((1−(x/(sinx)))/x^2 ) =−(1/6) .
$${we}\:{have}\:{sinx}\:=\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+{o}\left({x}^{\mathrm{3}} \right)\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\frac{{sinx}\:−{x}}{{x}^{\mathrm{2}} {sinx}}\:\:\sim\:\:\frac{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}}{{x}^{\mathrm{3}} }\:\rightarrow−\frac{\mathrm{1}}{\mathrm{6}}\:\:{when}\:{x}\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−\frac{{x}}{{sinx}}}{{x}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{6}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18
lim_(x→0)  ((sinx−x)/(x^2 sinx))  ((0/0))  lim_(x→0)  ((cosx−1)/(x^2 cosx+sinx.2x))((0/0))  lim_(x→0)  ((sinx)/(−x^2 sinx+cosx.2x+sinx.2+2xcosx)) ((0/0))  lim_(x→0)  (((sinx)/x)/(−xsinx+2cosx+2((sinx)/x)+2cosx))  =(1/(−0+2+2+2))=(1/6)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sinx}−{x}}{{x}^{\mathrm{2}} {sinx}}\:\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{cosx}−\mathrm{1}}{{x}^{\mathrm{2}} {cosx}+{sinx}.\mathrm{2}{x}}\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sinx}}{−{x}^{\mathrm{2}} {sinx}+{cosx}.\mathrm{2}{x}+{sinx}.\mathrm{2}+\mathrm{2}{xcosx}}\:\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{sinx}}{{x}}}{−{xsinx}+\mathrm{2}{cosx}+\mathrm{2}\frac{{sinx}}{{x}}+\mathrm{2}{cosx}} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{0}+\mathrm{2}+\mathrm{2}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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