Question Number 159744 by 0731619 last updated on 20/Nov/21
Answered by gsk2684 last updated on 20/Nov/21
$$\frac{\mathrm{0}}{\mathrm{0}}\:{form},\:{apply}\:{L}'{hospital}\:{rule} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{−\left(\mathrm{cos}\:\frac{\pi}{{x}}\right)\left(−\frac{\pi}{{x}^{\mathrm{2}} }\right)}{\left(\mathrm{sin}\:\pi{x}\right)\pi}=\underset{{x}\rightarrow\mathrm{2}} {\frac{\mathrm{1}}{\mathrm{4}}\mathrm{lim}\frac{\left(\mathrm{cos}\:\frac{\pi}{{x}}\right)}{\left(\mathrm{sin}\:\pi{x}\right)}\:} \\ $$$${again}\:{apply}\:{the}\:{same}\:{rule} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\left(−\mathrm{sin}\:\:\frac{\pi}{{x}}\right)\left(−\frac{\pi}{{x}^{\mathrm{2}} }\right)}{\left(\mathrm{cos}\:\:\pi{x}\right)\pi}\:=\frac{\mathrm{1}}{\mathrm{4}}\frac{\left(−\mathrm{sin}\:\:\frac{\pi}{\mathrm{2}}\right)\left(−\frac{\pi}{\mathrm{2}^{\mathrm{2}} }\right)}{\left(\mathrm{cos}\:\:\mathrm{2}\pi\right)\pi}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\frac{\left(−\mathrm{1}\right)\left(−\frac{\pi}{\mathrm{4}}\right)}{\left(\mathrm{1}\right)\pi}\:=\frac{\mathrm{1}}{\mathrm{16}} \\ $$