Question Number 146376 by iloveisrael last updated on 13/Jul/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:\mathrm{cos}\:\mathrm{3x}...\:\mathrm{cos}\:\mathrm{px}}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$
Answered by gsk2684 last updated on 13/Jul/21
$$\frac{\mathrm{0}}{\mathrm{0}}{form},\:{apply}\:{L}'{Hopital}\:{rule} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}.\mathrm{cos}\:\mathrm{2}{x}...\mathrm{cos}\:{px}+\mathrm{cos}\:{x}.\left(\mathrm{2sin}\:\mathrm{2}{x}\right)\mathrm{cos}\:\mathrm{3}{x}...\mathrm{cos}\:{px}+..\:}{\mathrm{2}{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+{p}^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}.\frac{{p}\left({p}+\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)}{\mathrm{6}}=\frac{{p}\left({p}+\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)}{\mathrm{12}} \\ $$
Commented by iloveisrael last updated on 13/Jul/21
$$\mathrm{wrong} \\ $$
Commented by gsk2684 last updated on 13/Jul/21
$${edited}.\:{see}\:{the}\:{answer} \\ $$
Commented by iloveisrael last updated on 13/Jul/21
$$\mathrm{yes}\:\mathrm{correct} \\ $$