How the following limit can be calculated:
$$\lim_{x \to 0} [(x+1)^x-1]^x$$ ?
I've estimated the $0^0$ limit by writing the function under the form $e^ {\ln\{...\}}$. Then, by applying twice l'Hospital rule, I've found the limit is 1.
I need help in order to obviate these large calculations.
Note that by Taylor's expansions
we have
$$(1+x)^x=e^{x\log(1+x)}=e^{x^2+o(x^2)}=1+x^2+o(x^2)$$
thus
$$[(x+1)^x-1] ^x=[1+x^2+o(x^2)-1] ^x=(x^2+o(x^2))^x=e^{x(\log x^2+o(x^2))}\to 1$$
indeed
$$x[\log x^2+o(x^2)]=x[\log x^2+\log (1+o(1))]=2x\log x+x\log(1+o(1))\to0+0=0$$
Compute first the limit of the log: $\;x\ln\bigl((x+1)^x-1\bigr)$.
We'll make an asymptotic expansion of $\;(x+1)^x$ near $0$: $$ (x+1)^x=\mathrm e^{x\ln(1+x)}=\mathrm e^{x(x+o(x))}=\mathrm e^{x^2+o(x^2)}=1+x^2+o(x^2)\quad\text{(by composition)}, $$
\begin{align} \text{so}\hspace6em x\ln\bigl((x+1)^x-1\bigr)&=x\ln\bigl(x^2+o(x^2)\bigr)=x\ln\bigl(x^2(1+o(1))\bigr)\qquad\qquad\\ &=2x\ln x+x\underbrace{\ln\bigl(1+o(1)\bigr)}_{\substack{\downarrow\\0}} \\ &=\underbrace{2x\ln x+o(x)}_{\substack{\downarrow\\0}}. \end{align} As the log tends to $0$, $\;\lim_{x\to 0}\bigl((x+1)^x-1\bigr)^x=1.$
Write your limit in the form $$\exp\left[\lim_{x\to 0}\left[x\ln((x+1)^x-1)\right]\right]$$
\displaystyle gives $$e^{\displaystyle \lim_{x\to 0}\ln((x+1)^x-1)x}$$ which feels 'unbalanced'. I think the better solution is to use \exp instead of e.
– Calvin Khor
Feb 03 '18 at 12:44