Computing Pointwise Limits How would you find the pointwise limit of 
$$\dfrac{\exp\left(\dfrac{x}{n}\right)}{n}$$
I'm confused on where to begin.
 A: If we're computing the pointwise limit, we would be evaluating
$$
\lim_{n\to\infty} \frac{e^{x/n}}{n}
$$
We could simply solve this as
$$
\begin{align}
\lim_{n\to\infty} \frac{e^{x/n}}{n} 
&= \left(\lim_{n\to\infty} e^{x/n}\right)\left(\lim_{n\to\infty} \frac{1}{n}\right)\\
&= (1)(0)\\
&= 0
\end{align}
$$
Thus, the pointwise limit of the function is zero.
We would then find that the series of functions $\{f_n\}_{n=1}^{\infty}$ given by $f_n(x)=\frac{e^{x/n}}{n}$ does not converge uniformly, since each $f_n$ is unbounded.

Proof that $f_n$ converges uniformly on $[0,1]$:
Take any $\epsilon>0$.  Choose $N\in\mathbb N$ so that $N>\dfrac{e}{\epsilon}$.  We note that for any $n>N$, for any $x\in[0,1]$
$$
\begin{align}
\left|f_n(x)-0\right|&=\left|f_n(x)\right|\\
&=\frac{e^{x/n}}{n}\\
&\leq \frac{e^{1/n}}{n}\\
&\leq \frac{e^1}{n}\\
&< \frac{e}{N}<\epsilon
\end{align}
$$
We conclude that $f_n\to 0$ uniformly on $[0,1]$.
A: Since this is a homework question, I am not going to solve it. I will provide the intuition.
For pointwise convergence, you want to show that for any given $x$, the function converges to zero. In this sense, you fix $x$, and show that the sequence converges. This is what the others have done.
For uniform convergence, there has to be an $n$ such that all values of the function are sufficiently close to the pointwise limit of the function.
To prove that it does not converge pointwise you just have to show that for any given $n$, there is some $x$ for which the value is of $$f(x,n)-lim_{n\rightarrow \infty} > m$$ where $m$ is fixed at some value. In this case for a given $n$, you can find x such that 
$$f(x,n)-0>1$$
For instance $n = 1$ 
\begin{align*}
(x,n) &= \frac{exp\left(\frac{x}{n}\right)}{n}\\
&= exp(x) \cdot exp\left(\frac{1}{n}\right) \frac{1}{n}\\
&= e^x e^1
\end{align*}
so any $x>-1$ is sufficient.
A: You can use the following theorem

Let $\{a_n\} \in \mathbb{N}$ and $\{ b_n \} \in \mathbb{N}$ be
  convergent sequences with $\lim_{n \to \infty} a_n = \alpha$ and
  $\lim_{n \to \infty} b_n = \beta$ Then
  $\{a_n b_n\}_n \in \mathbb{N}$ is convergent and
  $$\lim_{n \to \infty}(a_n b_n) = \alpha \beta$$

Taking $a_n = \exp(\frac{x}{n})$ and $b_n = \frac{1}{n}$.
Calculating the limit of each sequence $\lim_{n \to \infty} a_n = 1$ and $\lim_{n \to \infty}b_{n} = 0$ then applying the previous theorem to $\{ a_n \}$ and $\{ b_n \}$.
$$\lim_{n \to \infty} a_n b_n = (1)(0) = 0$$
