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Question-176721

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Question Number 176721 by youssefelaour last updated on 25/Sep/22
Answered by MJS_new last updated on 25/Sep/22
=lim_(x→(π/4)) ((−2sin x)/(2cos x)) =−1
$$=\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:{x}}{\mathrm{2cos}\:{x}}\:=−\mathrm{1} \\ $$
Answered by cortano1 last updated on 26/Sep/22
lim_(x→0) ((2cos (x+(π/4))−(√2))/(2sin (x+(π/4))−(√2))) = lim_(x→0) ((2{(1/2)(√2) cos x−(1/2)(√2) sin x}−(√2))/(2((1/2)(√2) sin x+(1/2)(√2) cos x)−(√2))) = lim_(x→0) (((√2) (cos x−1)−2(√2) sin (1/2)x cos (1/2)x)/( 2(√2) sin (1/2)x cos (1/2)x+(√2) (cos x−1))) =lim_(x→0) ((−2(√2) sin (1/2)x(sin (1/2)x+cos (1/2)x))/(2(√2) sin (1/2)x(cos (1/2)x−sin (1/2)x))) =lim_(x→0) ((sin (1/2)x+cos (1/2)x)/(sin (1/2)x−cos (1/2)x))= −1
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)−\sqrt{\mathrm{2}}}{\mathrm{2sin}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)−\sqrt{\mathrm{2}}}\: \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\right\}−\sqrt{\mathrm{2}}}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}\right)−\sqrt{\mathrm{2}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)−\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\left(\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)}{\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\left(\mathrm{cos}\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}=\:−\mathrm{1} \\ $$

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