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
$$=\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}{x}\mathrm{sin}\:\:{x}−\pi}{\mathrm{cos}\:{x}}=\left[\frac{\mathrm{0}}{\mathrm{0}}\right] \\ $$$$\overset{{L}'{Hopital}} {=}\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}{x}\mathrm{cos}\:{x}+\mathrm{2sin}\:{x}}{−{sinx}}= \\ $$$$=−\frac{\pi{cos}\left(\pi/\mathrm{2}\right)+\mathrm{2}{sin}\left(\pi/\mathrm{2}\right)}{{sin}\left(\pi/\mathrm{2}\right)}= \\ $$$$=−\mathrm{2} \\ $$
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
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{\mathrm{2xsin}\:\mathrm{x}−\pi}{\mathrm{cos}\:\mathrm{x}}\right)=\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\frac{\mathrm{2xsin}\:\mathrm{x}−\pi}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}\right) \\ $$$$\mathrm{let}\:\frac{\pi}{\mathrm{2}}−\mathrm{x}=\:\mathrm{t}\:,\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)−\pi}{\mathrm{sin}\:\mathrm{t}}\: \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\pi−\mathrm{2t}\right)\mathrm{cos}\:\mathrm{t}−\pi}{\mathrm{sin}\:\mathrm{t}} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\pi\left(\mathrm{cos}\:\mathrm{t}−\mathrm{1}\right)−\mathrm{2t}\:\mathrm{cos}\:\mathrm{t}}{\mathrm{sin}\:\mathrm{t}} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2}\pi\:\mathrm{sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}−\mathrm{2t}\:\mathrm{cos}\:\mathrm{t}}{\mathrm{sin}\:\mathrm{t}} \\ $$$$=−\mathrm{2}\: \\ $$
Commented by saly last updated on 13/Aug/21
$$\:\:{than}\:{you}\:{very}\:{much}\:! \\ $$