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Question Number 87061 by john santu last updated on 02/Apr/20
lim_(x→0) ((cos^3 (2x)−cos (x))/(cos^2 (2x)−cos (x))) =
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{2x}\right)−\mathrm{cos}\:\left(\mathrm{x}\right)}{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2x}\right)−\mathrm{cos}\:\left(\mathrm{x}\right)}\:=\: \\ $$
Commented by jagoll last updated on 02/Apr/20
i can try let cos 2x = t ⇒2cos^2 (x)−1=t cos (x) = (√((t+1)/2)) lim_(t→1) ((t^3 −(√((t+1)/2)))/(t^2 −(√((t+1)/2)))) = lim_(t→1) ((3t^2 −(1/( (√2))) ((1/(2(√(t+1))))))/(2t−(1/( (√2)))((1/(2(√(t+1))))))) = ((3−(1/4))/(2−(1/4))) = ((11)/7).
$$\mathrm{i}\:\mathrm{can}\:\mathrm{try} \\ $$$$\mathrm{let}\:\mathrm{cos}\:\mathrm{2x}\:=\:\mathrm{t}\:\Rightarrow\mathrm{2cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{1}=\mathrm{t} \\ $$$$\mathrm{cos}\:\left(\mathrm{x}\right)\:=\:\sqrt{\frac{\mathrm{t}+\mathrm{1}}{\mathrm{2}}} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{t}^{\mathrm{3}} −\sqrt{\frac{\mathrm{t}+\mathrm{1}}{\mathrm{2}}}}{\mathrm{t}^{\mathrm{2}} −\sqrt{\frac{\mathrm{t}+\mathrm{1}}{\mathrm{2}}}}\:=\: \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{3t}^{\mathrm{2}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{t}+\mathrm{1}}}\right)}{\mathrm{2t}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{t}+\mathrm{1}}}\right)}\:=\: \\ $$$$\frac{\mathrm{3}−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}}\:=\:\frac{\mathrm{11}}{\mathrm{7}}. \\ $$
Commented by john santu last updated on 02/Apr/20
waw ..cool...
$$\mathrm{waw}\:\:..\mathrm{cool}… \\ $$
Answered by MJS last updated on 02/Apr/20
(1) u(x)=cos^3 2x −cos x v(x)=cos^2 2x −cos x lim_(x→0) ((u(x))/(v(x))) =lim_(x→0) ((u′(x))/(v′(x))) =lim_(x→0) ((u′′(x))/(v′′(x))) =((11)/7) (2) let x=2arctan t; x=0 ⇔ t=0 lim_(x→0) ((cos^3 2x −cos x)/(cos^2 2x −cos x)) = =lim_(t→0) ((t^(10) −7t^8 +58t^6 −126t^4 +53t^2 −11)/(t^(10) −3t^8 +10t^6 +26t^4 +5t^2 −7)) =((11)/7)
$$\left(\mathrm{1}\right) \\ $$$${u}\left({x}\right)=\mathrm{cos}^{\mathrm{3}} \:\mathrm{2}{x}\:−\mathrm{cos}\:{x} \\ $$$${v}\left({x}\right)=\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\:−\mathrm{cos}\:{x} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}\left({x}\right)}{{v}\left({x}\right)}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}'\left({x}\right)}{{v}'\left({x}\right)}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}''\left({x}\right)}{{v}''\left({x}\right)}\:=\frac{\mathrm{11}}{\mathrm{7}} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{let}\:{x}=\mathrm{2arctan}\:{t};\:{x}=\mathrm{0}\:\Leftrightarrow\:{t}=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}^{\mathrm{3}} \:\mathrm{2}{x}\:−\mathrm{cos}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\:−\mathrm{cos}\:{x}}\:= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}^{\mathrm{10}} −\mathrm{7}{t}^{\mathrm{8}} +\mathrm{58}{t}^{\mathrm{6}} −\mathrm{126}{t}^{\mathrm{4}} +\mathrm{53}{t}^{\mathrm{2}} −\mathrm{11}}{{t}^{\mathrm{10}} −\mathrm{3}{t}^{\mathrm{8}} +\mathrm{10}{t}^{\mathrm{6}} +\mathrm{26}{t}^{\mathrm{4}} +\mathrm{5}{t}^{\mathrm{2}} −\mathrm{7}}\:=\frac{\mathrm{11}}{\mathrm{7}} \\ $$
Commented by jagoll last updated on 02/Apr/20
waw...funtastic
$$\mathrm{waw}…\mathrm{funtastic} \\ $$

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