Par définition, ${\displaystyle (1+\sin x)^{\frac {1}{x}}=\mathrm {e} ^{f(x)}}$ avec ${\displaystyle f(x)={\frac {\ln(1+\sin x)}{x}}\,{\underset {x\to 0}{\sim }}\,{\frac {\sin x}{x}}\,{\underset {x\to 0}{\to }}\,1}$, donc ${\displaystyle \lim _{x\to 0}{(1+\sin x)^{\frac {1}{x}}}=\mathrm {e} ^{1}=\mathrm {e} }$.

${\displaystyle (1+x)^{\frac {1}{x}}=\mathrm {e} ^{f(x)}}$ avec ${\displaystyle f(x)={\frac {\ln(1+x)}{x}}\,{\underset {x\to +\infty }{\sim }}\,{\frac {\ln(1+x)}{1+x}}\,{\underset {x\to +\infty }{\to }}\,0}$, donc ${\displaystyle \lim _{x\to +\infty }{(1+x)^{\frac {1}{x}}}=\mathrm {e} ^{0}=1}$.

${\displaystyle {\begin{aligned}{\sqrt[{3}]{x}}\left({\sqrt[{3}]{(x+1)^{2}}}-{\sqrt[{3}]{(x-1)^{2}}}\right)&=x^{\frac {1}{3}}\left(\left(x+1\right)^{\frac {2}{3}}-\left(x-1\right)^{\frac {2}{3}}\right)\\&=x\left(\left(1+{\frac {1}{x}}\right)^{\frac {2}{3}}-\left(1-{\frac {1}{x}}\right)^{\frac {2}{3}}\right)\\&\,{\underset {x\to +\infty }{\sim }}\,x\times {\frac {4}{3x}}\end{aligned}}}$

Donc

${\displaystyle \lim _{x\to +\infty }{\sqrt[{3}]{x}}\left({\sqrt[{3}]{(x+1)^{2}}}-{\sqrt[{3}]{(x-1)^{2}}}\right)={\frac {4}{3}}}$.

${\displaystyle \lim _{x\to (-1)^{+}}{\frac {{\sqrt {\pi }}-{\sqrt {\arccos }}x}{\sqrt {x+1}}}=\lim _{y\to 0^{+}}{\frac {{\sqrt {\pi }}-{\sqrt {\pi -y}}}{\sqrt {\cos(\pi -y)+1}}}={\sqrt {\pi }}\lim _{y\to 0^{+}}{\frac {1-{\sqrt {1-y/\pi }}}{\sqrt {1-\cos y}}}={\sqrt {\pi }}\lim _{y\to 0^{+}}{\frac {y/(2\pi )}{\sqrt {y^{2}/2}}}={\frac {1}{\sqrt {2\pi }}}}$.

Quand ${\displaystyle x\to +\infty }$, si ${\displaystyle b\leq a}$,

${\displaystyle \left(a^{x}+b^{x}\right)^{\frac {1}{x}}=a\left(1+\left({\frac {b}{a}}\right)^{x}\right)^{\frac {1}{x}}\to a}$

car pour tout ${\displaystyle x>0}$,

${\displaystyle 1<\left(1+\left({\frac {b}{a}}\right)^{x}\right)^{\frac {1}{x}}\leq 2^{\frac {1}{x}}}$ (et ${\displaystyle 2^{\frac {1}{x}}\to 2^{0}=1}$).

De même, si ${\displaystyle a\leq b}$, la limite est ${\displaystyle b}$.

Dans les deux cas, la limite est donc ${\displaystyle \max(a,b)}$.

Si ${\displaystyle b=a}$, la limite est évidemment ${\displaystyle a}$.

Supposons maintenant par exemple ${\displaystyle b<a}$.

${\displaystyle \left({\frac {a^{x}+b^{x}}{2}}\right)^{\frac {1}{x}}=a\left({\frac {1+\left({\frac {b}{a}}\right)^{x}}{2}}\right)^{\frac {1}{x}}}$.

Quand ${\displaystyle x\to 0}$,

${\displaystyle \ln \left({\frac {1+\left({\frac {b}{a}}\right)^{x}}{2}}\right)\sim {\frac {\left({\frac {b}{a}}\right)^{x}-1}{2}}\sim {\frac {x\ln {\frac {b}{a}}}{2}}}$

donc

${\displaystyle \left({\frac {a^{x}+b^{x}}{2}}\right)^{\frac {1}{x}}\to a\exp \left({\frac {\ln {\frac {b}{a}}}{2}}\right)=a{\sqrt {\frac {b}{a}}}={\sqrt {ab}}}$.

Dans tous les cas, la limite est donc ${\displaystyle {\sqrt {ab}}}$.

${\displaystyle (x^{2}+1)^{\alpha }-(x+1)^{2\alpha }=x^{2\alpha }\left[\left(1+{\frac {1}{x^{2}}}\right)^{\alpha }-\left(1+{\frac {1}{x}}\right)^{2\alpha }\right]=x^{2\alpha }\left[1+o\left({\frac {1}{x}}\right)-\left(1+{\frac {2\alpha }{x}}+o\left({\frac {1}{x}}\right)\right)\right]\,{\underset {x\to +\infty }{\sim }}\,-2\alpha x^{2\alpha -1}}$

donc la limite est :

${\displaystyle {\begin{cases}0&{\text{si }}\alpha <{\frac {1}{2}}\\-1&{\text{si }}\alpha ={\frac {1}{2}}\\-\infty &{\text{si }}\alpha >{\frac {1}{2}}.\end{cases}}}$

Si ${\displaystyle \alpha \leq 0}$, ${\displaystyle \lim _{x\to +\infty }{\left(1+{\frac {1}{x^{\alpha }}}\right)^{x}}=+\infty }$.

Si ${\displaystyle \alpha >0}$, ${\displaystyle \ln \left(\left(1+{\frac {1}{x^{\alpha }}}\right)^{x}\right)=x\ln \left(1+{\frac {1}{x^{\alpha }}}\right)\,{\underset {x\to +\infty }{\sim }}\,x^{1-\alpha }}$ donc :

• si ${\displaystyle 0<\alpha <1}$, ${\displaystyle \lim _{x\to +\infty }{\left(1+{\frac {1}{x^{\alpha }}}\right)^{x}}=\mathrm {e} ^{+\infty }={+\infty }}$ ;
• si ${\displaystyle \alpha >1}$, ${\displaystyle \lim _{x\to +\infty }{\left(1+{\frac {1}{x^{\alpha }}}\right)^{x}}=\mathrm {e} ^{0}=1}$ ;
• si ${\displaystyle \alpha =1}$, ${\displaystyle \lim _{x\to +\infty }{\left(1+{\frac {1}{x^{\alpha }}}\right)^{x}}=\mathrm {e} ^{1}=\mathrm {e} }$.

1. ${\displaystyle {\frac {\sinh u}{u}}\,{\underset {u\to +\infty }{\sim }}\,{\frac {\mathrm {e} ^{u}}{2u}}\,{\underset {u\to +\infty }{\to }}\,+\infty }$ et ${\displaystyle \operatorname {arcosh} v\,{\underset {v\to +\infty }{\sim }}\,\ln v}$

   donc

   ${\displaystyle \operatorname {arcosh} \left(x\sinh(1/x)\right)\,{\underset {x\to 0^{+}}{\sim }}\,\ln \left(x\sinh(1/x)\right)\,{\underset {x\to 0^{+}}{\sim }}\,\ln {\frac {x\mathrm {e} ^{1/x}}{2}}\,{\underset {x\to 0^{+}}{\sim }}\,{\frac {1}{x}}}$,

   c'est-à-dire

   ${\displaystyle \lim _{x\to 0^{+}}x\operatorname {arcosh} \left(x\sinh(1/x)\right)=1}$.

2. ${\displaystyle {\sqrt {x^{2}+x}}-{\sqrt {x^{2}-x}}=x\left({\sqrt {1+1/x}}-{\sqrt {1-1/x}}\right)\,{\underset {x\to +\infty }{\to }}\,1}$

   donc

   ${\displaystyle \mathrm {e} ^{\sqrt {x^{2}+x}}-\mathrm {e} ^{\sqrt {x^{2}-x}}=\mathrm {e} ^{\sqrt {x^{2}-x}}\left(\mathrm {e} ^{{\sqrt {x^{2}+x}}-{\sqrt {x^{2}-x}}}-1\right)\,{\underset {x\to +\infty }{\sim }}\,\mathrm {e} ^{\sqrt {x^{2}-x}}(\mathrm {e} -1)\,{\underset {x\to +\infty }{\to }}\,+\infty }$.

   Par conséquent, ${\displaystyle \mathrm {e} ^{-{\sqrt {x^{2}-x}}}-\mathrm {e} ^{-{\sqrt {x^{2}+x}}}\,{\underset {x\to +\infty }{=}}\,o\left(\mathrm {e} ^{\sqrt {x^{2}+x}}-\mathrm {e} ^{\sqrt {x^{2}-x}}\right)}$ et

   ${\displaystyle \sinh {\sqrt {x^{2}+x}}-\sinh {\sqrt {x^{2}-x}}={\frac {\mathrm {e} ^{\sqrt {x^{2}+x}}-\mathrm {e} ^{\sqrt {x^{2}-x}}+\mathrm {e} ^{-{\sqrt {x^{2}-x}}}-\mathrm {e} ^{-{\sqrt {x^{2}+x}}}}{2}}\,{\underset {x\to +\infty }{\to }}\,+\infty }$.

3. ${\displaystyle {\sqrt {x+1}}-{\sqrt {x}}={\sqrt {x}}\left({\sqrt {1+1/x}}-1\right)\,{\underset {x\to +\infty }{\sim }}\,{\frac {1}{2{\sqrt {x}}}}\,{\underset {x\to +\infty }{\to }}\,0}$

   donc

   ${\displaystyle \mathrm {e} ^{\sqrt {x+1}}-\mathrm {e} ^{\sqrt {x}}=\mathrm {e} ^{\sqrt {x}}\left(\mathrm {e} ^{{\sqrt {x+1}}-{\sqrt {x}}}-1\right)\,{\underset {x\to +\infty }{\sim }}\,{\frac {\mathrm {e} ^{\sqrt {x}}}{2{\sqrt {x}}}}\,{\underset {x\to +\infty }{\to }}\,+\infty }$.

   Par conséquent, ${\displaystyle \mathrm {e} ^{-{\sqrt {x}}}-\mathrm {e} ^{-{\sqrt {x+1}}}\,{\underset {x\to +\infty }{=}}\,o\left(\mathrm {e} ^{\sqrt {x+1}}-\mathrm {e} ^{\sqrt {x}}\right)}$ et

   ${\displaystyle \cosh {\sqrt {x+1}}-\cosh {\sqrt {x}}={\frac {\mathrm {e} ^{\sqrt {x+1}}-\mathrm {e} ^{\sqrt {x}}+\mathrm {e} ^{-{\sqrt {x}}}-\mathrm {e} ^{-{\sqrt {x+1}}}}{2}}\,{\underset {x\to +\infty }{\sim }}\,{\frac {\mathrm {e} ^{\sqrt {x}}}{4{\sqrt {x}}}}\,{\underset {x\to +\infty }{\to }}\,+\infty }$,

   si bien que

   ${\displaystyle \ln \left(\cosh {\sqrt {x+1}}-\cosh {\sqrt {x}}\right)\,{\underset {x\to +\infty }{\sim }}\,\ln \left({\frac {\mathrm {e} ^{\sqrt {x}}}{4{\sqrt {x}}}}\right)={\sqrt {x}}-\ln(4{\sqrt {x}})\,{\underset {x\to +\infty }{\sim }}\,{\sqrt {x}}}$

   et finalement,

   ${\displaystyle \lim _{x\to +\infty }\left(\cosh {\sqrt {x+1}}-\cosh {\sqrt {x}}\right)^{\frac {1}{\sqrt {x}}}=\lim _{x\to +\infty }\exp {\frac {\ln \left(\cosh {\sqrt {x+1}}-\cosh {\sqrt {x}}\right)}{\sqrt {x}}}=\exp 1=\mathrm {e} }$.

1. ${\displaystyle \lim _{x\to 1^{-}}{\frac {\arcsin x}{\tan {\frac {\pi x}{2}}}}={\frac {\pi /2}{+\infty }}=0}$.
2. ${\displaystyle \lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{\frac {x+1}{x}}=1^{1}=1}$.
3. ${\displaystyle {\frac {1}{\sin x}}\ln {\frac {\sin x}{x}}\,{\underset {x\to 0}{\sim }}\,{\frac {1}{x}}\left({\frac {\sin x}{x}}-1\right)\,{\underset {x\to 0}{\sim }}\,-{\frac {x}{6}}}$ donc ${\displaystyle \lim _{x\to 0}\left({\frac {\sin x}{x}}\right)^{\frac {1}{\sin x}}=\mathrm {e} ^{0}=1}$.
4. ${\displaystyle \operatorname {cotan} bx\ln(\cos ax)\,{\underset {x\to 0}{\sim }}\,{\frac {\cos ax-1}{bx}}\,{\underset {x\to 0}{\sim }}\,{\frac {-a^{2}x}{2b}}}$ donc ${\displaystyle \lim _{x\to 0}(\cos ax)^{\operatorname {cotan} bx}=\mathrm {e} ^{0}=1}$.
5. ${\displaystyle \ln(1+x)-\ln(1-x)\,{\underset {x\to 0}{\sim }}\,x-(-x)=2x}$ et ${\displaystyle \sin x\,{\underset {x\to 0}{\sim }}\,x}$ donc ${\displaystyle \lim _{x\to 0}\left({\frac {1+x}{1-x}}\right)^{1/\sin x}=\mathrm {e} ^{2}}$.