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Question Number 115889 by bemath last updated on 29/Sep/20

lim_(x→0)  ((((1+cos 2x))^(1/(3 ))  −(2)^(1/(3 )) )/(x^2 .sin 3x))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{2}}}{{x}^{\mathrm{2}} .\mathrm{sin}\:\mathrm{3}{x}} \\ $$

Answered by bemath last updated on 29/Sep/20

 lim_(x→0)  ((((1+(1−((4x^2 )/2))))^(1/(3 )) −(2)^(1/(3 )) )/(x^2 .sin 3x)) =  lim_(x→0)  ((((2−2x^2 ))^(1/(3 )) −(2)^(1/(3 )) )/(x^2 .sin 3x)) =  lim_(x→0)  (((2)^(1/(3 ))  (((1−x^2 ))^(1/(3 )) −1 ))/(3x^3 )) =   lim_(x→0)  (((2)^(1/(3 ))  (1−(x^2 /3)−1))/(3x^3 )) =   lim_(x→0)  ((−(2)^(1/(3 ))  x^2 )/(9x^3 )) = lim_(x→0)  ((−(2)^(1/(3 )) )/x) = ±∞

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}\:}]{\mathrm{2}}}{{x}^{\mathrm{2}} .\mathrm{sin}\:\mathrm{3}{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}−\mathrm{2}{x}^{\mathrm{2}} }−\sqrt[{\mathrm{3}\:}]{\mathrm{2}}}{{x}^{\mathrm{2}} .\mathrm{sin}\:\mathrm{3}{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}\:\left(\sqrt[{\mathrm{3}\:}]{\mathrm{1}−{x}^{\mathrm{2}} }−\mathrm{1}\:\right)}{\mathrm{3}{x}^{\mathrm{3}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}\:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{3}{x}^{\mathrm{3}} }\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\sqrt[{\mathrm{3}\:}]{\mathrm{2}}\:{x}^{\mathrm{2}} }{\mathrm{9}{x}^{\mathrm{3}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\sqrt[{\mathrm{3}\:}]{\mathrm{2}}}{{x}}\:=\:\pm\infty \\ $$

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