Question Number 211241 by efronzo1 last updated on 01/Sep/24
$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{x}^{\mathrm{2}} −\mathrm{cos}\:\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:\right)}{\mathrm{x}^{\mathrm{6}} }\:=? \\ $$
Commented by Frix last updated on 01/Sep/24
$$\mathrm{Using}\:\mathrm{Taylor}\:\mathrm{polynomials}\:\mathrm{I}\:\mathrm{get}\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by klipto last updated on 01/Sep/24
$$\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{taylor}}\:\boldsymbol{\mathrm{swift}}\:\boldsymbol{\mathrm{lol}}… \\ $$
Commented by Frix last updated on 01/Sep/24
$$\mathrm{I}\:\mathrm{know}\:\mathrm{a}\:\mathrm{swift}\:\mathrm{tailor}\:\mathrm{too}. \\ $$
Commented by York12 last updated on 03/Sep/24
$$\mathrm{lol} \\ $$
Answered by mehdee1342 last updated on 01/Sep/24
$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{cosx}^{\mathrm{2}} −\mathrm{1}−{cos}\left({sin}^{\mathrm{2}} {x}\right)+\mathrm{1}}{{x}^{\mathrm{6}} } \\ $$$$={li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{4}} {x}}{{x}^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{sin}^{\mathrm{4}} {x}−{x}^{\mathrm{4}} }{{x}^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\left({sinx}−{x}\right)\left({sinx}+{x}\right)\left({sin}^{\mathrm{2}} {x}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\left(−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \right)\left(\mathrm{2}{x}\right)\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{6}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\:\checkmark \\ $$$$ \\ $$