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Question-195947




Question Number 195947 by cortano12 last updated on 13/Aug/23
Answered by MM42 last updated on 14/Aug/23
((2cosa)/(sin^3 a))
$$\frac{\mathrm{2}{cosa}}{{sin}^{\mathrm{3}} {a}} \\ $$
Answered by qaz last updated on 14/Aug/23
lim_(x→0) ((cot a+(((cot a)′)/(1!))(2x)+(((cot a)′′)/(2!))(2x)^2 −2(cot a+(((cot a)′)/(1!))x+(((cot a)′′)/(2!))x^2 )+cot a+o(x^2 ))/x^2 )  =lim_(x→0) (((cot a)′′x^2 +o(x^2 ))/x^2 )=2cot a∙csc^2  a
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{cot}\:{a}+\frac{\left(\mathrm{cot}\:{a}\right)'}{\mathrm{1}!}\left(\mathrm{2}{x}\right)+\frac{\left(\mathrm{cot}\:{a}\right)''}{\mathrm{2}!}\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{cot}\:{a}+\frac{\left(\mathrm{cot}\:{a}\right)'}{\mathrm{1}!}{x}+\frac{\left(\mathrm{cot}\:{a}\right)''}{\mathrm{2}!}{x}^{\mathrm{2}} \right)+\mathrm{cot}\:{a}+{o}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{cot}\:{a}\right)''{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }=\mathrm{2cot}\:{a}\centerdot{csc}^{\mathrm{2}} \:{a} \\ $$

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