Menu Close

lim-x-0-cos-3-8x-1-6x-2-




Question Number 104838 by bramlex last updated on 24/Jul/20
lim_(x→0)  ((cos ^3 (8x)−1)/(6x^2 )) ?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\:^{\mathrm{3}} \left(\mathrm{8}{x}\right)−\mathrm{1}}{\mathrm{6}{x}^{\mathrm{2}} }\:? \\ $$
Answered by john santu last updated on 24/Jul/20
lim_(x→0)  (((cos 8x−1)(cos^2 8x+cos 8x+1))/(6x^2 ))  lim_(x→0) ((−2sin^2 4x)/(6x^2 )) × 3 = −((16×3)/3)  = −16  (JS ⊛)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:\mathrm{8}{x}−\mathrm{1}\right)\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{8}{x}+\mathrm{cos}\:\mathrm{8}{x}+\mathrm{1}\right)}{\mathrm{6}{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{4}{x}}{\mathrm{6}{x}^{\mathrm{2}} }\:×\:\mathrm{3}\:=\:−\frac{\mathrm{16}×\mathrm{3}}{\mathrm{3}} \\ $$$$=\:−\mathrm{16}\:\:\left({JS}\:\circledast\right)\: \\ $$$$ \\ $$
Commented by bramlex last updated on 24/Jul/20
♠⧫
$$\spadesuit\blacklozenge \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *