Uniform convergence of function sequence Given the function sequence below: 

$$f_n(x)=e^{-x^2/n}, \quad x \in \Bbb R$$

Is the functional sequence $f_n(x)$ convergent uniformly?
I know that $f_n(x)$ is convergent to $f(x)=1$ point wise, but I cannot show uniformly convergent of this.
 A: I hope you know that if a sequence of continuous function converges uniformly, the the following holds: 
\begin{align} \lim\limits_{x\to a}  \lim\limits_{n\to \infty}f_n(x)=\lim\limits_{{n\to \infty}}   \lim\limits_{x\to a}f_n(x)\end{align}
If that does not hold, then the function does not converge uniformly. Now look at this:
\begin{align} \lim\limits_{x\to \infty}  \lim\limits_{n\to \infty}f_n(x) = \lim\limits_{x\to \infty}  1 =1\end{align}
But on the other hand we have:
\begin{align} \lim\limits_{{n\to \infty}}   \lim\limits_{x\to \infty}f_n(x) =\lim\limits_{{n\to \infty}}   \lim\limits_{x\to \infty} e^{\frac{-x^2}{n}} =\lim\limits_{{n\to \infty}} 0 = 0 \neq 1 = \lim\limits_{x\to \infty}  \lim\limits_{n\to \infty}f_n(x) \end{align}
Thus, the sequence doesn't converge uniformly.
A: This sequence of functions is not uniformly convergent.
To see this, suppose the sequence $f_n(x)$ is uniformly convergent. Then it would have to converge to $1$, since uniform convergence implies pointwise convergence, and you have identified that the sequence converges pointwise to $1$.
And if $f_n(x)$ uniformly converges to $1$, then by definition, for every $\epsilon > 0$, there exists an $N \in \mathbb N $ such that
$$ n > N \implies \sup_{x \in \mathbb R} | e^{-x^2/n} - 1 | < \epsilon.$$
Are you able to determine $\sup_{x \in \mathbb R} | e^{-x^2/n} - 1 | $, for a given $n$?
Can you find an $\epsilon$ such that the above inequality fails? Good luck!
