较复杂的幂指极限

青青子衿

\begin{align*}
\lim_{x\to+\infty}\dfrac{\left[\dfrac{x}{2}\ln\left(\dfrac{x+1}{x-1}\right)\right]^{x^4}}{\exp\!\left(\dfrac{x^2}{3}\right)}=\exp\!\left(\dfrac{13}{90}\right)
\end{align*}

\begin{align*}
\lim_{x\to+\infty}\frac{\left(\sqrt{x (x+1)}-\sqrt{x (x-1)}\right)^{x^4}}{\exp \left(\frac{x^2}{8}\right)}=\exp\left(\dfrac{3}{64}\right)\\
\end{align*}

Czhang271828

左式对数为
\[
\lim_{x\to +\infty}x^4\ln\left(\dfrac{x}{2}\ln \dfrac{x+1}{x-1}\right)-\dfrac{x^2}{3}.
\]
换元 $t=\dfrac{1}{x}$, 上式为(下省略 $\lim_{t\to 0^+}$)
\[
t^{-4}\cdot \ln\left[\dfrac{1}{2t}\ln \left(1+\dfrac{2t}{1-t}\right)\right]-\dfrac{1}{3t^2}.
\]
Taylor 展开最内侧项, 得
\[
t^{-4}\cdot \ln\left[\dfrac{1}{2t}\cdot \left(2t + \dfrac{2 t^3}3 + \dfrac{2 t^5}5 + O(t^6)\right)\right]-\dfrac{1}{3t^2}.
\]
即
\[
t^{-4}\cdot \ln\left(1 + \dfrac{t^2}3 + \dfrac{t^4}5 + O(t^5)\right)-\dfrac{1}{3t^2}.
\]
再展开至 $O(t^5)$ 项, 得
\[
t^{-4}\cdot \left(\dfrac{t^2}3+\dfrac{13t^4}{90}+O(t^5)\right)-\dfrac{1}{3t^2}.
\]
从而原极限为 $e^{13/90}$.

hbghlyj

$\tanh(x)=\frac12\ln\left(1+x\over1-x\right)$

青青子衿

hbghlyj 发表于 2023-3-26 16:54
$\tanh(x)=\frac12\ln\left(1+x\over1-x\right)$

\begin{align*}
L&=\lim_{x\to+\infty}\dfrac{\left[\dfrac{x}{2}\ln\left(\dfrac{x+1}{x-1}\right)\right]^{x^4}}{\exp\!\left(\dfrac{x^2}{3}\right)}\\
&=\exp\!\left(\lim_{x\to+\infty}\left(x^4\ln\left(\dfrac{x}{2}\ln\left(\dfrac{x+1}{x-1}\right)\right)-\dfrac{x^2}{3}\right)\right)\\
&=\exp\!\left(\lim_{\,\,t\to0^{+}}\left(\dfrac{1}{t^4}\ln\left(\dfrac{1}{2t}\ln\left(\dfrac{1+t}{1-t}\right)\right)-\dfrac{1}{3t^2}\right)\right)\\
&=\exp\!\left(\lim_{\,\,t\to0^{+}}\dfrac{1}{t^2}\left(\dfrac{1}{t^2}\ln\left(\dfrac{1}{2t}\ln\left(\dfrac{1+t}{1-t}\right)\right)-\dfrac{1}{3}\right)\right)\\
&=\exp\!\left(\lim_{\,\,t\to0^{+}}\dfrac{1}{t^2}\left(\dfrac{1}{t^2}\ln\left(1+\dfrac{t^2}{3}+\dfrac{t^4}{5}\right)-\dfrac{1}{3}\right)\right)\\

&=\exp\!\left(\lim_{\,\,t\to0^{+}}\dfrac{1}{t^2}\left(\dfrac{1}{t^2}\left(\dfrac{t^2}{3}+\dfrac{t^4}{5}-\dfrac{1}{2}\left(\dfrac{t^2}{3}+\dfrac{t^4}{5}\right)^2\right)-\dfrac{1}{3}\right)\right)\\
&=\exp\!\left(\lim_{\,\,t\to0^{+}}\dfrac{1}{t^2}\left(\dfrac{1}{t^2}\left(\frac{t^2}{3}+\frac{13 t^4}{90}-\frac{t^6}{15}-\frac{t^8}{50}\right)-\dfrac{1}{3}\right)\right)\\
&=\exp\left(\dfrac{13}{90}\right)
\end{align*}