求[lim _{x ightarrow 0} frac{e^{(1+x)^{1/x}}-(1+x)^{e / x}}{x^{2}}]
由泰勒公式可知
egin{align*}
(1+x)^{1 / x} &=exp left{frac{ln (1+x)}{x}
ight}=exp left{frac{x-frac{x^{2}}{2}+frac{x^{3}}{3}+oleft(x^{4}
ight)}{x}
ight} \ &=e cdot exp left{-frac{x}{2}+frac{x^{2}}{3}+oleft(x^{3}
ight)
ight}\
&=eleft[1+left(-frac{x}{2}+frac{x^{2}}{3}+oleft(x^{3}
ight)
ight)+frac{left(-frac{x}{2}+frac{x^{2}}{3}+oleft(x^{3}
ight)
ight)^{2}}{2}+oleft(x^{3}
ight)
ight] \
&=e-frac{e}{2} x+frac{11 e}{24} x^{2}+oleft(x^{3}
ight).
end{align*}
于是
[e^{(1+x)^{1/{x}}}=e^{e}-frac{1}{2} e^{e+1} x+frac{1}{24} e^{e+1}(3 e+11) x^{2}+oleft(x^{2}
ight).]
类似地有
[(1+x)^{e / x}=e^{e}-frac{1}{2} e^{e+1} x+frac{1}{24} e^{e+1}(3 e+8) x^{2}+oleft(x^{2}
ight).]
于是所求极限为
[lim _{x
ightarrow 0} frac{e^{(1+x)^{l x}}-(1+x)^{e / x}}{x^{2}}=frac{1}{8} e^{e+1}.]
若$lim_{n oinfty}a_n=b>0$,判断$sum_ {n=1}^{infty}left(frac{b}{a_n} ight)^n$的敛散性.
取$a_n=bsqrt[n]{n^2} o b$,则$sum_ {n=1}^{infty}left(frac{b}{a_n} ight)^n=sum_ {n=1}^{infty}frac{1}{n^2}$收敛.
取$a_n=bsqrt[n]{n} o b$,则$sum_ {n=1}^{infty}left(frac{b}{a_n}
ight)^n=sum_ {n=1}^{infty}frac{1}{n}$发散.
求极限$lim_{n oinfty}n^2left(arctanfrac{a}{n}-arctan{a}{n+1} ight)$.