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Question Number 160995 by bobhans last updated on 10/Dec/21

lim_(x→0) (((sin 2x−2tan x)^2 +(1−cos 2x)^3 )/(tan^7 6x +sin^6 x))=?

$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{sin}\:\mathrm{2x}−\mathrm{2tan}\:\mathrm{x}\right)^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{3}} }{\mathrm{tan}\:^{\mathrm{7}} \mathrm{6x}\:+\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}}=? \\ $$

Commented by blackmamba last updated on 11/Dec/21

(•)(sin 2x−2tan x)^2 = (2sin xcos x−((2sin x)/(cos x)))^2 = (((2sin x cos^2 x−2sin x)/(cos x)))^2 = ((4sin^2 x)/(cos^2 x))(cos^2 x−1)^2 = ((4sin^6 x)/(cos^2 x)) (•) (1−cos 2x)^3 = (2sin^2 x)^3 =8sin^6 x L = lim_(x→0) ((4sin^6 x+8sin^6 x cos^2 x)/(cos^2 x(tan^7 6x+sin^6 x))) L = lim_(x→0) ((2cos^2 x+1)/(cos^2 x)) × lim_(x→0) ((4sin^6 x)/(tan^7 6x+sin^6 x)) L= 3×4×lim_(x→0) (x^6 /(x^6 (6^7 x+1))) = 12×lim_(x→0) (1/(1+6^7 x)) L = 12×1=12

$$\:\:\left(\bullet\right)\left(\mathrm{sin}\:\mathrm{2}{x}−\mathrm{2tan}\:{x}\right)^{\mathrm{2}} =\:\left(\mathrm{2sin}\:{x}\mathrm{cos}\:{x}−\frac{\mathrm{2sin}\:{x}}{\mathrm{cos}\:{x}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:=\:\left(\frac{\mathrm{2sin}\:{x}\:\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{2sin}\:{x}}{\mathrm{cos}\:{x}}\right)^{\mathrm{2}} =\:\frac{\mathrm{4sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\left(\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:=\:\frac{\mathrm{4sin}\:^{\mathrm{6}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$$\:\:\left(\bullet\right)\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)^{\mathrm{3}} =\:\left(\mathrm{2sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{3}} =\mathrm{8sin}\:^{\mathrm{6}} {x} \\ $$$$\:\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4sin}\:^{\mathrm{6}} {x}+\mathrm{8sin}\:^{\mathrm{6}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{tan}\:^{\mathrm{7}} \mathrm{6}{x}+\mathrm{sin}\:^{\mathrm{6}} {x}\right)} \\ $$$$\:\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4sin}\:^{\mathrm{6}} {x}}{\mathrm{tan}\:^{\mathrm{7}} \mathrm{6}{x}+\mathrm{sin}\:^{\mathrm{6}} {x}} \\ $$$$\:\:\mathcal{L}=\:\mathrm{3}×\mathrm{4}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{6}} }{{x}^{\mathrm{6}} \left(\mathrm{6}^{\mathrm{7}} {x}+\mathrm{1}\right)}\:=\:\mathrm{12}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{6}^{\mathrm{7}} {x}} \\ $$$$\:\:\mathcal{L}\:=\:\mathrm{12}×\mathrm{1}=\mathrm{12} \\ $$

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