Question Number 112782 by bemath last updated on 09/Sep/20
$$\mathrm{without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{x}−\pi} \\ $$
Answered by john santu last updated on 09/Sep/20
$${setting}\:{x}\:=\:\pi+{s}\:,\:{s}\rightarrow\mathrm{0} \\ $$$$\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\frac{{s}}{\mathrm{2}}\right)}{{s}}\:=\:\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:\left(\frac{{s}}{\mathrm{2}}\right)}{{s}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Sep/20
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)}{\left(\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by Aziztisffola last updated on 09/Sep/20
$$\underset{\mathrm{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{x}−\pi}=\mathrm{cos}'\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mid_{\mathrm{x}=\pi} =−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by malwan last updated on 09/Sep/20
$${this}\:{is}\:{L}^{'} {Hopital}\: \\ $$
Commented by Aziztisffola last updated on 09/Sep/20
$$\mathrm{I}\:\mathrm{use}\:\mathrm{this}\:\mathrm{property}\:;\mathrm{derivitive}\:\mathrm{in}\:\mathrm{one} \\ $$$$\mathrm{point}\:{x}_{\mathrm{0}\:} ;\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{l}'\mathrm{Hopital}\:\mathrm{rule} \\ $$$$\underset{{x}\rightarrow{x}_{\mathrm{0}} } {\mathrm{lim}}\frac{{f}\left({x}\right)−{f}\left({x}_{\mathrm{0}} \right)}{{x}−{x}_{\mathrm{0}} }={f}\:'\left({x}_{\mathrm{0}} \right) \\ $$$$\:\mathrm{let}\:{f}\left({x}\right)=\mathrm{cos}\left(\frac{{x}}{\mathrm{2}}\right)\:\:{and}\:{x}_{\mathrm{0}} =\pi \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{f}\left({x}\right)−{f}\left(\pi\right)}{{x}−\pi}\:={f}\:'\left(\pi\right)=\:\mathrm{cos}'\left(\frac{\pi}{\mathrm{2}}\right) \\ $$