Question Number 82682 by zainal tanjung last updated on 23/Feb/20
$$\mathrm{Help}\:\mathrm{me}\:\mathrm{please}....!! \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{Lim}}\:\left(\frac{\mathrm{1}}{\mathrm{ex}}\right)^{\mathrm{6x}} =... \\ $$
Commented by john santu last updated on 23/Feb/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{ex}}−\mathrm{1}\right)^{\mathrm{6}{x}} \:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}−{ex}}{{ex}}\right)^{\mathrm{6}{x}} \\ $$$$={e}\:^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−{ex}}{{ex}}\right)×\mathrm{6}{x}} \:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{6}{x}−\mathrm{6}{ex}^{\mathrm{2}} }{{ex}}\right)} \\ $$$$=\:{e}\:^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−{ex}}{{e}}\right)} =\:{e}\:^{\frac{\mathrm{1}}{{e}}} \:.\: \\ $$
Commented by zainal tanjung last updated on 23/Feb/20
![lim_(m→∞) [1+((m−e)/e)]^(6/m) lim_(m→∞) [(1+(1/(((e/(m−e))))))^([(e/(m−e))]) ]^([((m−e)/e)]((6/m))) e^(lim_(m→∞) [((6m−6e)/(me))]) =e^(6/e) =((e^6 ))^(1/e)](Q82725.png)
$$\underset{\mathrm{m}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{1}+\frac{\mathrm{m}−\mathrm{e}}{\mathrm{e}}\right]^{\frac{\mathrm{6}}{\mathrm{m}}} \\ $$$$\underset{\mathrm{m}\rightarrow\infty} {\mathrm{lim}}\:\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{e}}{\mathrm{m}−\mathrm{e}}\right)}\right)^{\left[\frac{\mathrm{e}}{\mathrm{m}−\mathrm{e}}\right]} \right]^{\left[\frac{\mathrm{m}−\mathrm{e}}{\mathrm{e}}\right]\left(\frac{\mathrm{6}}{\mathrm{m}}\right)} \\ $$$$\mathrm{e}^{\underset{\mathrm{m}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{\mathrm{6m}−\mathrm{6e}}{\mathrm{me}}\right]} =\mathrm{e}^{\frac{\mathrm{6}}{\mathrm{e}}} =\sqrt[{\mathrm{e}}]{\mathrm{e}^{\mathrm{6}} \:} \\ $$$$ \\ $$
Commented by john santu last updated on 24/Feb/20
$${m}?\:{where}\:{you}\:{got}\:{m}? \\ $$