Showing $\dfrac{e^{-nx}}{n} < \epsilon$ I am trying to prove that $f_n(x)=\dfrac{e^{-nx}}{n}\rightarrow f(x)=0$ uniformly on $x\in[0, \infty)$. To do this I am trying to show that: $$\dfrac{e^{-nx}}{n} &lt; \epsilon$$
I tried to do it this way, but I am not sure if it is correct:
\begin{align}
&amp;\dfrac{e^{-nx}}{n} \\ 
&amp;&lt;\dfrac{1}{n} \quad \forall n&gt;N_1 \ \exists N_1\in \mathbb{N} \\
&amp;&lt;\epsilon \quad \forall n&gt;N_2 \ \exists N_2\in \mathbb{N} \\ \\
\therefore \dfrac{e^{-nx}}{n}&amp;&lt;\epsilon \quad \forall n&gt;N=\max\{N_1, N_2\}
\end{align}

This in turn implies uniform convergence. Is this the correct way to do it?
Let $n \geq 1$ then $g(x)=e^{-nx}$ is a decreasing function defiend on $[0,+\infty)$

Thus $e^{-nx} \leq e^0=1$ thus $$\sup_{x \geq 0}|\frac{e^{-nx}}{n}| \leq \frac{1}{n}$$

Thus  is $\epsilon &gt;0$ then for $n \geq n_0=[\frac{1}{\epsilon}]+1$ we have that $$ \sup_{x \geq 0}|\frac{e^{-nx}}{n}| \leq \epsilon$$
You got the idea.

Need to show that for $\epsilon &gt;0$ there is a 

$N \in \mathbb{Z^+}$ such that for $x \in [0,\infty)$ and  

$n \gt N$ we have $|\dfrac{e^{-nx}}{n} -0| \lt \epsilon.$

$\dfrac{e^{-nx}}{n} =\dfrac{1}{ne^{nx}} =$

$\dfrac{1}{n(1+nx+(nx)^2/2!+...)} \le \dfrac{1}{n}.$

Archimedes :

There is a $N \in \mathbb{Z^+}$ such that

$N &gt;1/\epsilon.$

For $x  \in [0,\infty )$ and $n &gt;N$ we have

$\dfrac{e^{-nx}}{n} \le 1/n \lt 1/N \lt \epsilon.$