I'm having some trouble evaluating the $\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}$ in the context of real analysis. Denote this sequence by $(a_k)$
If so, then I need to prove that for all $\varepsilon>0$
$$\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right| <\varepsilon$$
We know that $\lim_{x\to\infty}1/x=0$, however $|1/x|\leq\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}\right|$, so I'm not sure this information is useful. I considered the sequential characterization of continuity, so if the sequence above converges to 0, then $f((a_k))$ converges to $f(0)$, but there is also not an obvious choice of $f$. Any suggestions into how to evaluate the limit?
We will show that $$\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}=0.$$ We let $\varepsilon >0$ be given. We need to find a $\delta >0$ such that $|x-0|<\delta$ implies ${\displaystyle \left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right|<\varepsilon }$.
In this post, a user proved that $\left(1+\frac{1}{x}\right)^{x}<3.$ Also for every $x>1$, $1/x<x$. Choose $\delta=\varepsilon/3$. $${\displaystyle \left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right|<\frac{3}{x}<3x<3\delta<\varepsilon }.$$