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