Distribution in $n$-dimensional ball From an $n$-dimensional ball of radius $\sqrt{n}$, randomly choose a point $(x_1, .., x_n)$.  Prove that with $n \to \infty$, the random variable $x_1$ tends to a random variable with a normal distribution.
 A: We can write down the marginal pdf of $X_1^{(n)}$:
$$
f_{X_1^{(n)}}(x)=\begin{cases}
\dfrac{V_{n-1}(\sqrt{n-x^2})}{V_n(\sqrt{n})} & x^2<n\\
0 & \text{otherwise}
\end{cases}
$$
where $V_n(R)$ is the volume of the ball of radius $R$ in Euclidean $n$-space
$$
V_n(R)=\frac{\pi^{n/2}}{\Gamma(\frac{n}2+1)}R^n.
$$
So for $n>x^2$,
\begin{align*}
f_{X_1^{(n)}}(x)&=\frac{\dfrac{\pi^{(n-1)/2}}{\Gamma(\frac{n+1}2)}(n-x^2)^{(n-1)/2}}{\dfrac{\pi^{n/2}}{\Gamma(\frac{n}2+1)}\sqrt{n}^n}\\
&=\underbrace{\left[\dfrac{\Gamma(\frac{n}2+1)}{\sqrt{n\pi}\Gamma(\frac{n+1}2)}\right]}_{\to\frac1{\sqrt{2\pi}}}\underbrace{\left(1-\frac{x^2}{n}\right)^{(n-1)/2}}_{\to\exp(-x^2/2)}.\\
&\to\frac1{\sqrt{2\pi}}\exp(-x^2/2)
\end{align*}
Hence $X_1^{(n)}\xrightarrow{D}N(0,1)$ as $n\to\infty$.
A: Slightly less computational than user10354138's answer, but with the same calculus inderpinnings:
You can write your $x=(x_1,\ldots,x_n)$ as $\sqrt n R_n  Z/\| Z\|_2$, where $Z=(Z_1,\ldots,Z_n)\sim N(0,I_n)$ is a standard Gaussian vector in $\mathbb R^n$ and where $R_n$, independent of $Z$,  has density function $n r^{n-1}$ on $[0,1]$.  
The idea here is that $Z/\|Z\|_2$  is uniformly distributed on the unit sphere in $n$-space, that $R_n Z/\|Z\|_2$ is uniformly distributed in the unit ball, and the factor $\sqrt n$ is what is needed to give the distributional limit. The cumulative distribution function for $R_n$ is $F_R(r)=P(R_n\le r) = r^n$, for $0\le r \le 1$, since the ratio of the volume of the radius $r$ ball to the radius $1$ ball is $r^n$, in $n$-space.  The density of $R_n$ is the derivative of this, $nr^{n-1}$ on $[0,1]$.
So the first component, $x_1$ has the same distribution as $R_n Z_1/ \sqrt{\|Z\|_2^2/n}.$  The ratio $\|Z\|_2^2/n=\sum_i^n Z_i^2/n$   tends , by the weak law of large numbers, to $1$ in probability as $n\to\infty$.  And $R_n$ tends to $1$ in probability, too.  By Slutsky's theorem, then, the distribution of $R_n Z_1/ \sqrt{\|Z\|_2^2/n}$ tends to that of $Z_1$, that is, to $N(0,1)$.
