(a) We compute the $L^{2}$ distance as

$$

d_{2}\left(x_{n},(0,0)\right)=\sqrt{\left(\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)\right)+\left(\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)\right)^{2}}=\frac{1}{n}

$$

(b) Since the distance is $1 / n$ and $1 / n \rightarrow 0$ in $\mathbb{R}$, then $x_{n} \rightarrow(0,0)$.

(c) We compute the $L^{1}$ distance as

$$

\begin{array}{l}

d_{1}\left(x_{n},(0,0)\right)=\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)-0\right|+\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)-0\right|= \\

\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)\right|+\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)\right|= \\

\left|\frac{1}{n}\right|\left(\left|\cos \left(\frac{n \pi}{2}\right)\right|+\left|\sin \left(\frac{n \pi}{2}\right)\right|\right)

\end{array}

$$

Notice that $\cos (n \pi / 2)$ and $\sin (n \pi / 2)$ take on the values $-1,0$, or $1 .$ In particular, when one is 0, the other is $\pm 1$ and vice versa. Thus, in absolute values, exactly one is 0 and one is $1 .$ So, our distance reduces to

$$

d_{1}\left(x_{n},(0,0)\right)=1 / n .

$$

(d) From (c), $d_{1}\left(x_{n},(0,0)\right)=1 / n \rightarrow 0 .$ Thus, $x_{n} \rightarrow(0,0)$ in the $L^{1}$ metric.

(e) We compute the $L^{\infty}$ distance as

$$

\begin{array}{c}

d_{\infty}\left(x_{n},(0,0)\right)=\max \left\{\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)-0\right|,\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)-0\right|\right\}= \\

\max \left\{\left|\frac{1}{n} \cos \left(\frac{n \pi}{2}\right)\right|,\left|\frac{1}{n} \sin \left(\frac{n \pi}{2}\right)\right|\right\}= \\

\left|\frac{1}{n}\right| \max \left\{\left|\cos \left(\frac{n \pi}{2}\right)\right|,\left|\sin \left(\frac{n \pi}{2}\right)\right|\right\}

\end{array}

$$

As mentioned above, the two terms are either 0 or 1 at opposite times. Thus, the maximum is equal to

1. So, the distance is given by

$$

d_{\infty}\left(x_{n},(0,0)\right)=\frac{1}{n} \text { . }

$$

(f) From (e), we have that $d_{\infty}\left(x_{n},(0,0)\right)=\frac{1}{n} \rightarrow 0 .$ Thus, $x_{n} \rightarrow(0,0)$.

(g) Notice that for all $n,|\cos (n \pi / 2)| \neq|\sin (n \pi / 2)| .$ Thus, in particular, they are never both 0 at the same time. So, they are never equal to $(0,0) .$ So, $d_{d}\left(x_{n},(0,0)\right)=1$ for all $n$. This does not converge to 0, so $x_{n} \not A(0,0)$ (or any other point).