Sequence of functions that converges to Gaussian Define $f_n:[0,\infty)\to\mathbb{R}$ as follows:
$$
f_n(x) = \begin{cases} \left(1-\frac{x^2}{n}\right)^n & 0\leq x\leq \sqrt{n} \\ 0 & \text{otherwise}\end{cases}
$$
I need to show that $f_n$ converges to $e^{-x^2}$ uniformly.
Pointwise convergence is easy since the limit of $\left(1-\frac{x^2}{n}\right)^n$ is $e^{-x^2}$. I tried to evaluate the distance
$$
\left|e^{-x^2} - \left(1-\frac{x^2}{n}\right)^n\right|
$$
but I was stuck since I remembered that the convergence of $\left(1-\frac{x^2}{n}\right)^n$ to $e^{-x^2}$ depends on $x$, that is the point after which the elements of the sequence gets arbitrarily close depends on the choice of $x$
 A: Hint: write $e^{-x^2}$ as $\left(e^{-x^2/n}\right)^n $ and exploit the inequality:
$$ \left|x^n-y^n\right| \leq n\left|x-y\right|\cdot\max(x,y)^{n-1} $$
together with the Taylor expansion of $e^{-z}-(1-z) = O(z^2)$ in a neighbourhood of the origin.
A: I would start with proving that 
$$
g_n(x) = \begin{cases} \left(1-\frac{x}{n}\right)^n & 0\leq x\leq n \\ 0 & \text{otherwise}\end{cases}
$$ converges uniformly to $g(x) = e^{-x}$ for $x \geq 0$.  Then we will use
$$\sup\limits_{0\leq x\leq \sqrt{n}} |g_n(x^2) - g(x)| = \sup\limits_{0\leq x\leq n} |g_n(x) - g(x)|.$$
Let's examine $g(x) - g_n(x)$ for $x \in [0,n]$:


*

*$g(0) - g_n(0) = 0$.

*$g$ is a solution of $g^\prime = -g$.

*$g_n$ is a solution of $g_n^\prime = -g_n^{1-\frac{1}{n}}$.  Let's write it as $$g_n^\prime = -g_n + (g_n - g_n^{1-\frac{1}{n}}).$$ Let us denote the parenthesized term $-h(x)$. 

*By the mean value theorem:
$$ -h(x) = \frac{1}{n}g_n^{1-\xi}\log g_n, \qquad 0< \xi < \frac{1}{n}.$$ However, the function $x\mapsto x^{1-\xi}\log x$ for $0 < x \leq 1$ is bounded by $0$ from above and by $- e^{-1}/1-\xi$ from below. Thus,
$$ 0 \geq -h(x) \geq - \frac{e^{-1}}{n(1-\xi)} \geq  - \frac{e^{-1}}{n - 1}.$$

*Subtract the equations for $g$ and $g_n$:
$$ (g - g_n)^\prime = - (g - g_n) + h,\quad 0 \geq h(x) \geq \frac{e^{-1}}{n - 1}.$$ Obviously, $g(0) = g_n(0)$.
When you express the solution and plug in the bounds for $h(x)$, you will obtain $$0 \leq g(x) - g_n(x) \leq \frac{e^{-1}}{n - 1}.$$

*Estimate the $g - g_n$ for $x > n$ is easy. The conclusion is $g - g_n \rightrightarrows 0$.

I haven't shown that the equation for $g_n$ is solvable in some sense. However, as $g_n$ is a polynomial, a check that $g_n$ and $h$ posess the desired properties should be an easy excersise.
