Question Number 115195 by bemath last updated on 24/Sep/20
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{8}}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{4}{x}}{\left(\pi−\mathrm{8}{x}\right)^{\mathrm{3}} }\:?\: \\ $$
Answered by bobhans last updated on 24/Sep/20
$${let}\:{x}\:=\:\frac{\pi}{\mathrm{8}}+{p}\:;\:{p}\rightarrow\mathrm{0} \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\left(\mathrm{4}{p}+\frac{\pi}{\mathrm{2}}\right)−\mathrm{cos}\:\left(\mathrm{4}{p}+\frac{\pi}{\mathrm{2}}\right)}{\left(−\mathrm{8}{p}\right)^{\mathrm{3}} }\:= \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{tan}\:\mathrm{4}{p}+\mathrm{sin}\:\mathrm{4}{p}}{−\mathrm{512}{p}^{\mathrm{3}} }\:=\: \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{4}{p}−\mathrm{sin}\:\mathrm{4}{p}}{\mathrm{512}{p}^{\mathrm{3}} }\:=\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{4}{p}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}{p}\right)}{\mathrm{512}{p}^{\mathrm{3}} } \\ $$$$=\:\frac{\mathrm{4}×\mathrm{8}}{\mathrm{64}×\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$