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Question-98823




Question Number 98823 by john santu last updated on 16/Jun/20
Commented by bramlex last updated on 16/Jun/20
L′Hopital   lim_(x→0) ((sin x+xcos x)/(3x^2 )) = lim_(x→0) ((cos x+cos x−xsin x)/(6x))  lim_(x→0)  ((2cos x−xsin x)/(6x)) = ∞ ?
$${L}'{Hopital}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}+{x}\mathrm{cos}\:{x}}{\mathrm{3}{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}+\mathrm{cos}\:{x}−{x}\mathrm{sin}\:{x}}{\mathrm{6}{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}−{x}\mathrm{sin}\:{x}}{\mathrm{6}{x}}\:=\:\infty\:?\: \\ $$
Commented by bemath last updated on 16/Jun/20
lim_(x→0)  ((x(x−(1/3)x^3 +o(x^3 )))/x^3 ) =   lim_(x→0)  ((x−(1/3)x^3 +o(x^3 ))/x^2 ) = ∞
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({x}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{3}} \right)\right)}{{x}^{\mathrm{3}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} }\:=\:\infty \\ $$

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