Question Number 160501 by LEKOUMA last updated on 30/Nov/21
$${Calculate} \\ $$$$\left.\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{2}}\right){x}}{\mathrm{1}−\sqrt{{x}}} \\ $$$$\left.\mathrm{2}\right)\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{e}^{\mathrm{1}+{x}} }{\left(\mathrm{1}+{x}\right)^{{x}} }−\frac{{x}}{{e}} \\ $$
Commented by cortano last updated on 01/Dec/21
$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\mathrm{x}\right)}{\mathrm{1}−\sqrt{\mathrm{x}}}\: \\ $$$$\:\mathrm{let}\:\mathrm{1}−\sqrt{\mathrm{x}}\:=\:\mathrm{u}\:;\:\mathrm{x}=\left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{2}} \\ $$$$\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{2}} \right)}{\mathrm{u}} \\ $$$$\:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{2u}\right)}{\mathrm{u}} \\ $$$$\:=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\pi}{\mathrm{2}}\mathrm{u}\left(\mathrm{u}−\mathrm{2}\right)}{\mathrm{u}}\:=\:−\pi \\ $$
Answered by qaz last updated on 01/Dec/21
$$\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\mathrm{x}}{\mathrm{1}−\sqrt{\mathrm{x}}}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\mathrm{x}}{−\mathrm{ln}\sqrt{\mathrm{x}}}=\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{−\frac{\pi}{\mathrm{2}}\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\mathrm{x}}{−\frac{\mathrm{1}}{\mathrm{2x}}}=\pi \\ $$