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Question-10398




Question Number 10398 by amir last updated on 07/Feb/17
Answered by arge last updated on 07/Feb/17
por l′hopital,    y= (1/x^2 ) − (1/(tan^2 x))    y=x^(−2)   − ((cos^2 x)/(sen^2 x))    y′= −2x^(−3) −((sen^2 x(−2senxcosx)−cos^2 x(2senxcosx))/(sen^4 x))    y′= −2x^(−3)  − (A/((1−cos^2 x)^2 ))    y′=0/∞=0
$${por}\:{l}'{hopital}, \\ $$$$ \\ $$$${y}=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {x}} \\ $$$$ \\ $$$${y}={x}^{−\mathrm{2}} \:\:−\:\frac{{cos}^{\mathrm{2}} {x}}{{sen}^{\mathrm{2}} {x}} \\ $$$$ \\ $$$${y}'=\:−\mathrm{2}{x}^{−\mathrm{3}} −\frac{{sen}^{\mathrm{2}} {x}\left(−\mathrm{2}{senxcosx}\right)−{cos}^{\mathrm{2}} {x}\left(\mathrm{2}{senxcosx}\right)}{{sen}^{\mathrm{4}} {x}} \\ $$$$ \\ $$$${y}'=\:−\mathrm{2}{x}^{−\mathrm{3}} \:−\:\frac{{A}}{\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${y}'=\mathrm{0}/\infty=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Answered by mrW1 last updated on 09/Feb/17
lim_(x→0) ((1/x^2 )−(1/(tan^2  x)))  =lim_(x→0) [((1/x)+((cos x)/(sin x)))((1/x)−((cos x)/(sin x)))]  =lim_(x→0) [((sin x+xcos x)/(xsin x))×((sin x−xcos x)/(xsin x))]  =lim_(x→0) [((sin x+xcos x)/(xsin^2  x))×((sin x−xcos x)/x)]  =lim_(x→0) [(((((sin )/x))+cos x)/((((sin x)/x))^2 ))×((sin x−xcos x)/x^3 )]  =lim_(x→0) (((((sin )/x))+cos x)/((((sin x)/x))^2 ))×lim_(x→0) ((sin x−xcos x)/x^3 )  =((1+1)/1^2 )×lim_(x→0) ((sin x−xcos x)/x^3 )  =2×lim_(x→0) ((sin x−xcos x)/x^3 )  =2×lim_(x→0) ((cos x−cos x+xsin x)/(3x^2 ))       ← applying L′hopital′s rule: lim_(x→c)   ((f(x))/(g(x)))=lim_(x→c)  ((f′(x))/(g′(x)))  =2×lim_(x→0) ((xsin x)/(3x^2 ))        =2×lim_(x→0) ((sin x)/(3x))  =(2/3)×lim_(x→0) ((sin x)/x)  =(2/3)×1  =(2/3)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:{x}}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right)\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right)\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{sin}\:{x}+{x}\mathrm{cos}\:{x}}{{x}\mathrm{sin}\:{x}}×\frac{\mathrm{sin}\:{x}−{x}\mathrm{cos}\:{x}}{{x}\mathrm{sin}\:{x}}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{sin}\:{x}+{x}\mathrm{cos}\:{x}}{{x}\mathrm{sin}^{\mathrm{2}} \:{x}}×\frac{\mathrm{sin}\:{x}−{x}\mathrm{cos}\:{x}}{{x}}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\left(\frac{\mathrm{sin}\:}{{x}}\right)+\mathrm{cos}\:{x}}{\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} }×\frac{\mathrm{sin}\:{x}−{x}\mathrm{cos}\:{x}}{{x}^{\mathrm{3}} }\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{sin}\:}{{x}}\right)+\mathrm{cos}\:{x}}{\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} }×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}−{x}\mathrm{cos}\:{x}}{{x}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}+\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}−{x}\mathrm{cos}\:{x}}{{x}^{\mathrm{3}} } \\ $$$$=\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}−{x}\mathrm{cos}\:{x}}{{x}^{\mathrm{3}} } \\ $$$$=\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}−\mathrm{cos}\:{x}+{x}\mathrm{sin}\:{x}}{\mathrm{3}{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\leftarrow\:{applying}\:{L}'{hopital}'{s}\:{rule}:\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}=\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:\frac{{f}'\left({x}\right)}{{g}'\left({x}\right)} \\ $$$$=\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{sin}\:{x}}{\mathrm{3}{x}^{\mathrm{2}} }\:\:\:\:\:\: \\ $$$$=\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}}{\mathrm{3}{x}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}}{{x}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{1} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by amir last updated on 09/Feb/17
thank you dear mrW1.
$${thank}\:{you}\:{dear}\:{mrW}\mathrm{1}. \\ $$

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