We know that
$$\left(1 + \frac{1}{x}\right)^x < e < \left(1 + \frac{1}{x}\right)^{x+1}$$ for all $x\in\mathbb{R}^{>0}$. My question is what is the infimum of $\delta\in\mathbb{R}^{>0}$ such that $$\left(1 + \frac{1}{x}\right)^{x+\delta}>e$$ for all $x\in\mathbb{R}^{>0}$?
$$\inf\{\delta\in\Bbb R\mid\forall x>0\quad(1+1/x)^{x+\delta}>e\}=\frac12$$ because: $$\begin{align}(\forall x>0\quad(1+1/x)^{x+\delta}>e)&\iff(\forall y>0\quad(\delta+1/y)\ln(1+y)>1)\\ &\iff(\forall t>1\quad\delta>\frac1{\ln t}-\frac1{t-1})\\ &\iff(\forall h>0\quad\delta>f(h):=\frac{e^h-1-h}{h(e^h-1)}), \end{align}$$ $\lim_0f=\frac12,$ and $$\forall h>0\quad e^h-1-h=\sum_{n\ge2}\frac{h^n}{n!}<\sum_{n\ge2}\frac{h^n}{2((n-1)!)}=\frac{h(e^h-1)}2.$$