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Question Number 193563 by SaRahAli last updated on 16/Jun/23
Answered by Gamil last updated on 16/Jun/23
Answered by Subhi last updated on 16/Jun/23
lim_(x→(π/4)) ((sin^3 (x)−cos^3 (x).((π/4)−x))/(sin((π/4)−x)((π/4)−x))) lim_(x→(π/4)) ((sin^3 (x)−cos^3 (x))/(((π/4)−x))) ⇛(((sin(x)−cos(x))(((sin(x)+cos(x))^2 −sin(x)cos(x)))/(((π/4)−x))) lim_(x→(π/4)) ((−cos(2x)(sin(x)+cos(x)))/(((π/4)−x)))+(((cos(x)−sin(x)sin(2x))/(2((π/4)−x))).((sin(x)+cos(x))/(sin(x)+cos(x))) lim_(x→(π/4)) ((−sin(2((π/4)−x))(sin(x)+cos(x)))/(((π/4)−x)))+((sin(4((π/4)−x))/(4((π/4)−x).(sin(x)+cos(x)))) lim_(x→(π/4)) −2(sin(x)+cos(x))+(1/(sin(x)+cos(x)))=−2(√2)+((√2)/2)=((−3(√2))/2)
$$ \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \frac{{sin}^{\mathrm{3}} \left({x}\right)−{cos}^{\mathrm{3}} \left({x}\right).\left(\frac{\pi}{\mathrm{4}}−{x}\right)}{{sin}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\left(\frac{\pi}{\mathrm{4}}−{x}\right)} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \frac{{sin}^{\mathrm{3}} \left({x}\right)−{cos}^{\mathrm{3}} \left({x}\right)}{\left(\frac{\pi}{\mathrm{4}}−{x}\right)}\:\Rrightarrow\frac{\left({sin}\left({x}\right)−{cos}\left({x}\right)\right)\left(\left(\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\mathrm{2}} −{sin}\left({x}\right){cos}\left({x}\right)\right)\right.}{\left(\frac{\pi}{\mathrm{4}}−{x}\right)} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \frac{−{cos}\left(\mathrm{2}{x}\right)\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)}{\left(\frac{\pi}{\mathrm{4}}−{x}\right)}+\frac{\left({cos}\left({x}\right)−{sin}\left({x}\right){sin}\left(\mathrm{2}{x}\right)\right.}{\mathrm{2}\left(\frac{\pi}{\mathrm{4}}−{x}\right)}.\frac{{sin}\left({x}\right)+{cos}\left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}\: \\ $$$$\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} \frac{−{sin}\left(\mathrm{2}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right)\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)}{\left(\frac{\pi}{\mathrm{4}}−{x}\right)}+\frac{{sin}\left(\mathrm{4}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right.}{\mathrm{4}\left(\frac{\pi}{\mathrm{4}}−{x}\right).\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}} −\mathrm{2}\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)+\frac{\mathrm{1}}{{sin}\left({x}\right)+{cos}\left({x}\right)}=−\mathrm{2}\sqrt{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{−\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$
Answered by mnjuly1970 last updated on 16/Jun/23
lim_( x→(π/4)) ((sin^( 3) (x)−cos^( 3) (x))/(sin((π/4) −x))) =? lim_( x→(π/4)) ((−(√2) sin((π/4) −x)(1+(1/2) sin(2x)))/(sin((π/(4 )) −x))) = −(√2) . (1+(1/2) sin((π/2)))=−((3(√2))/2) hint : sin(x)−cos(x)=(√2) sin(x−(π/4)) sin^( 3) (x)−cos^( 3) (x)=(sin(x)−cos(x))(1+(1/2) sin(2x)) ■
$$ \\ $$$$\:\:\:\:{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\frac{{sin}^{\:\mathrm{3}} \left({x}\right)−{cos}^{\:\mathrm{3}} \left({x}\right)}{{sin}\left(\frac{\pi}{\mathrm{4}}\:−{x}\right)}\:=? \\ $$$$\:\:\:\:\:\:{lim}_{\:{x}\rightarrow\frac{\pi}{\mathrm{4}}} \:\frac{−\sqrt{\mathrm{2}}\:{sin}\left(\frac{\pi}{\mathrm{4}}\:−{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\left(\mathrm{2}{x}\right)\right)}{{sin}\left(\frac{\pi}{\mathrm{4}\:}\:−{x}\right)} \\ $$$$\:\:\:\:\:\:=\:−\sqrt{\mathrm{2}}\:.\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\left(\frac{\pi}{\mathrm{2}}\right)\right)=−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\:\:\:{hint}\::\:\:{sin}\left({x}\right)−{cos}\left({x}\right)=\sqrt{\mathrm{2}}\:{sin}\left({x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:{sin}^{\:\mathrm{3}} \left({x}\right)−{cos}^{\:\mathrm{3}} \left({x}\right)=\left({sin}\left({x}\right)−{cos}\left({x}\right)\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\left(\mathrm{2}{x}\right)\right) \\ $$$$\:\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$

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