Question Number 187270 by MikeH last updated on 15/Feb/23
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}\:+\:\mathrm{sin}\left(\mathrm{2}{x}\right)\:+\mathrm{1}}{{x}^{\mathrm{2}} −\pi^{\mathrm{2}} }\:= \\ $$$$\mathrm{A}.\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\:\:\:\:\:\:\:\:{C}.\:\mathrm{1} \\ $$$${B}.\:\frac{\mathrm{1}}{\pi}\:\:\:\:\:\:\:\:\:{D}.\:\mathrm{Does}\:\mathrm{Not}\:\mathrm{exist} \\ $$
Answered by horsebrand11 last updated on 15/Feb/23
$$\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\pi−{x}\right)−\mathrm{sin}\:\mathrm{2}\left(\pi−{x}\right)}{\left({x}+\pi\right)\left({x}−\pi\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}−\mathrm{sin}\:\mathrm{2}{x}}{−{x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:\left[\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}}\:−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{{x}}\:\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\pi}\:\left[\:\mathrm{0}−\mathrm{2}\:\right]=\frac{\mathrm{1}}{\pi} \\ $$
Commented by MikeH last updated on 16/Feb/23
$$\mathrm{thanks}\: \\ $$
Answered by Frix last updated on 15/Feb/23