Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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

♠⧫

$$\spadesuit\blacklozenge \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com