Question-150467

Question Number 150467 by saly last updated on 12/Aug/21

Commented by saly last updated on 12/Aug/21

$$\:\:{Do}\:{you}\:{help}\:{me} \\ $$

Answered by aleks041103 last updated on 13/Aug/21

=lim_(x→π/2) ((2xsin x−π)/(cos x))=[(0/0)] =^(L′Hopital) lim_(x→π/2) ((2xcos x+2sin x)/(−sinx))= =−((πcos(π/2)+2sin(π/2))/(sin(π/2)))= =−2

$$=\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}{x}\mathrm{sin}\:\:{x}−\pi}{\mathrm{cos}\:{x}}=\left[\frac{\mathrm{0}}{\mathrm{0}}\right] \\ $$$$\overset{{L}'{Hopital}} {=}\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}{x}\mathrm{cos}\:{x}+\mathrm{2sin}\:{x}}{−{sinx}}= \\ $$$$=−\frac{\pi{cos}\left(\pi/\mathrm{2}\right)+\mathrm{2}{sin}\left(\pi/\mathrm{2}\right)}{{sin}\left(\pi/\mathrm{2}\right)}= \\ $$$$=−\mathrm{2} \\ $$

Commented by saly last updated on 13/Aug/21

$$\:{thank}\:{you}\:{very}\:{much} \\ $$

Commented by EDWIN88 last updated on 13/Aug/21

$${false}.\:\frac{{d}}{{dx}}\left(\mathrm{2}{x}\:\mathrm{sin}\:{x}\right)=\mathrm{2sin}\:{x}+\mathrm{2}{x}\mathrm{cos}\:{x} \\ $$

Commented by aleks041103 last updated on 13/Aug/21

$${Yes},\:{you}'{re}\:{right}!\:{I}'{ll}\:{correct}\:{it}\:{right} \\ $$$${now}. \\ $$

Answered by liberty last updated on 13/Aug/21

lim_(x→(π/2)) (((2xsin x−π)/(cos x)))=lim_(x→(π/2)) (((2xsin x−π)/(sin ((π/2)−x)))) let (π/2)−x= t ,t→0 lim_(t→0) ((2((π/2)−t)sin ((π/2)−t)−π)/(sin t)) =lim_(t→0) (((π−2t)cos t−π)/(sin t)) =lim_(t→0) ((π(cos t−1)−2t cos t)/(sin t)) =lim_(t→0) ((−2π sin^2 (1/2)t−2t cos t)/(sin t)) =−2

$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{\mathrm{2xsin}\:\mathrm{x}−\pi}{\mathrm{cos}\:\mathrm{x}}\right)=\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\frac{\mathrm{2xsin}\:\mathrm{x}−\pi}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}\right) \\ $$$$\mathrm{let}\:\frac{\pi}{\mathrm{2}}−\mathrm{x}=\:\mathrm{t}\:,\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)−\pi}{\mathrm{sin}\:\mathrm{t}}\: \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\pi−\mathrm{2t}\right)\mathrm{cos}\:\mathrm{t}−\pi}{\mathrm{sin}\:\mathrm{t}} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\pi\left(\mathrm{cos}\:\mathrm{t}−\mathrm{1}\right)−\mathrm{2t}\:\mathrm{cos}\:\mathrm{t}}{\mathrm{sin}\:\mathrm{t}} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2}\pi\:\mathrm{sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}−\mathrm{2t}\:\mathrm{cos}\:\mathrm{t}}{\mathrm{sin}\:\mathrm{t}} \\ $$$$=−\mathrm{2}\: \\ $$

Commented by saly last updated on 13/Aug/21

$$\:\:{than}\:{you}\:{very}\:{much}\:! \\ $$

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