Question Number 160900 by LEKOUMA last updated on 08/Dec/21
$${Calculate} \\ $$$$\left.\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right)^{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{a}}\right)^{\frac{\mathrm{1}}{{x}−{a}}} \\ $$
Commented by cortano last updated on 09/Dec/21
$$\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:=\mathrm{e}^{\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{sin}\:\mathrm{a}}{\mathrm{x}−\mathrm{a}}} =\:\mathrm{e}^{\mathrm{cos}\:\mathrm{a}} \\ $$
Commented by stelor last updated on 11/Dec/21
![1) 1 2) l=lim_(x→a) e^((ln (sinx)−ln(sina))/(x−a)) ln(l)=[ln(sinx)]_a ′=cotan (a) l=e^(cotan(a))](https://www.tinkutara.com/question/Q161074.png)
$$ \\ $$$$\left.\mathrm{1}\right)\:\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{l}=\underset{{x}\rightarrow{a}} {\mathrm{lim}}{e}^{\frac{\mathrm{ln}\:\left({sinx}\right)−{ln}\left({sina}\right)}{{x}−{a}}} \:\: \\ $$$$\:\:{ln}\left({l}\right)=\left[{ln}\left({sinx}\right)\right]_{{a}} '={co}\mathrm{tan}\:\left({a}\right) \\ $$$${l}={e}^{{cotan}\left({a}\right)} \\ $$