How to prove that the convergence of the sequence $f_n(x) = \Big(1+\frac{1}{nx}\Big)^{nx}$ is not uniform.

Question

Let $f_n(x) = \Big(1+\frac{1}{nx}\Big)^{nx}$ for $x\in (0, 1) $.

Obviously the sequence converges pointwise and the limit function is: $$\lim_{n \to \infty}f_n(x) = \lim_{n \to \infty}\Big(1+\frac{1}{nx}\Big)^{nx} = \lim_{t \to \infty}\Big(1+\frac{1}{t}\Big)^{t} = e = f(x).$$ However, convergence is not uniform, and one of the ways to prove that is by using the fact that $f_n(\frac{1}{n}) = 2$, but I don't know how to put it all together. How to justify using $x=\frac{1}{n}$ as a counter example while trying to (dis)prove $\displaystyle\lim_{n \to \infty }{\sup_{x\in (0, 1)}{\big|f_n(x)-e\big|}}=0$ ?

How does using $f_n(\frac{1}{n}) = 2$ prove that above mentioned supremum does not converge to $0$?

Please, help.

Answer 1

$$\sup_{x\in (0,1)} |f_n(x) - e| \geq \left| f_n\left(\frac 1n\right) - e\right| = e-2$$ is bounded away from 0.