Question-132007

Question Number 132007 by abdullahquwatan last updated on 10/Feb/21

Answered by bramlexs22 last updated on 10/Feb/21

lim_(x→0) ((cos 2x+sin 3x(cos x−cos 2x)−cos x)/(sin^2 x+cos 2x−1)) = lim_(x→0) (((cos 2x−cos x)(1−sin 3x))/(cos 2x−cos^2 x)) = lim_(x→0) (((2cos^2 x−cos x−1))/(cos^2 x−1 )) ×lim_(x→0) (1−sin 3x) = 1 × lim_(x→0) (((2cos x+1)(cos x−1))/((cos x+1)(cos x−1))) = (3/2)

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

Commented by abdullahquwatan last updated on 10/Feb/21

$$\mathrm{thanks} \\ $$

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Question-132007

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