Question Number 10374 by ridwan balatif last updated on 06/Feb/17 | ||
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)\mathrm{sin}\left(\mathrm{3x}−\mathrm{3}\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2}\left(\mathrm{1}−\mathrm{sin2x}\right)}=...? \\ $$ | ||
Answered by mrW1 last updated on 06/Feb/17 | ||
$${let}\:{u}={x}−\frac{\pi}{\mathrm{4}} \\ $$$${with}\:{x}\rightarrow\frac{\pi}{\mathrm{4}},\:{u}\rightarrow\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}=\mathrm{sin}\:\left(\mathrm{2}{u}+\frac{\pi}{\mathrm{2}}\right)=\mathrm{cos}\:\left(\mathrm{2}{u}\right)=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{u} \\ $$$$ \\ $$$${L}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)\mathrm{sin}\left(\mathrm{3x}−\mathrm{3}\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2}\left(\mathrm{1}−\mathrm{sin2x}\right)} \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{u}\:\mathrm{sin}\:\left(\mathrm{3}{u}\right)}{\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{u}}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{u}×\left(\frac{\mathrm{sin}\:\mathrm{3}{u}}{\mathrm{3}{u}}\right)×\mathrm{3}{u}}{\mathrm{4}×\left(\frac{\mathrm{sin}\:{u}}{{u}}\right)^{\mathrm{2}} ×{u}^{\mathrm{2}} } \\ $$$$=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}×\left(\frac{\mathrm{sin}\:\mathrm{3}{u}}{\mathrm{3}{u}}\right)}{\mathrm{4}×\left(\frac{\mathrm{sin}\:{u}}{{u}}\right)^{\mathrm{2}} }=\frac{\mathrm{3}×\mathrm{1}}{\mathrm{4}×\mathrm{1}^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}} \\ $$ | ||
Commented by ridwan balatif last updated on 06/Feb/17 | ||
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Answered by arge last updated on 08/Feb/17 | ||
$${por}\:{l}'{hopital}, \\ $$$$ \\ $$$${y}=\frac{\left({x}−\frac{{x}}{\mathrm{4}}\right){sen}\left(\mathrm{3}{x}−\frac{\mathrm{12}+\pi}{\mathrm{4}}\right)}{\mathrm{2}\left(\mathrm{1}−{sen}\mathrm{2}{x}\right)} \\ $$$$ \\ $$$${y}'=\frac{\mathrm{2}\left(\mathrm{1}−{sen}\mathrm{2}{x}\right)\left[\left({x}−\frac{{x}}{\mathrm{4}}\right){cos}\left(\mathrm{3}{x}−\frac{\mathrm{12}+\pi}{\mathrm{4}}\right)+{sen}\left(\mathrm{3}{x}−\frac{\mathrm{12}+\pi}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\right]−\left({x}−\frac{{x}}{\mathrm{4}}\right){sen}\left(\mathrm{3}{x}−\frac{\mathrm{12}+\pi}{\mathrm{4}}\right)×\mathrm{2}\left(−{cos}\mathrm{2}{x}\right)}{\mathrm{4}\left(\mathrm{1}−\mathrm{2}{sen}\mathrm{2}{x}+\mathrm{4}{sen}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\right)} \\ $$$$ \\ $$$${y}'=\frac{{A}}{\mathrm{0}} \\ $$$$ \\ $$$${A}\underset{} {=}\:\:−\mathrm{134}.\mathrm{65},\:{y}'=\infty\because\because\because\because\because{Rta} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ | ||
Answered by bahmanfeshki last updated on 02/Mar/17 | ||
$$\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}=\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} =\mathrm{2sin}^{\mathrm{2}} \:\left({x}+\frac{\pi}{\mathrm{4}}\right) \\ $$$${x}+\frac{\pi}{\mathrm{4}}={t} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}\mathrm{sin}\:\mathrm{3}{t}}{\mathrm{4sin}\:^{\mathrm{2}} {t}}=\frac{\mathrm{3}}{\mathrm{4}}\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{sin}\:\mathrm{3}{t}}{\mathrm{3}{t}}}{\left(\frac{\mathrm{sin}\:{t}}{{t}}\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}} \\ $$ | ||