Q:

Prove: $limX_n=0 \implies lim\left|X_n\right|^k = 0, k>0$. What if $k<0$?

This is what I've done so far. Not sure it it's correct tough, and what in case when $k<0$. 
*Proof:
Since $X_n$ converges to 0, by $\epsilon$ definition, we have:
$$(\forall\epsilon>0)(\exists n_0\in \mathbb{N})(\forall n \in \mathbb{N})(n\ge n_0 \implies \left|X_n\right| < \epsilon)$$
Since this holds $\forall\epsilon > 0$, it also must hold for $\epsilon<1$, so almost every element of $X_n$ is smaller than 1. In other words $(\exists n_1 \in \mathbb{N})(\forall n \in \mathbb{N})(n \ge n_1 \implies \left|X_n\right| < 1)$. Because $\left|X_n\right| < 1$, we can write it down like $\frac{1}{m}, m > 1$, so: $$\left|X_n\right| = \frac{1}{m} > \frac{1}{m^k} = \left|X_n\right|^k, k>0$$Plugin back in $\epsilon$, we have:
$$(\forall\epsilon>0)(\forall n \in N)(n \ge max(n_0, n_1) \implies \left|X_n\right|^k < \left|X_n\right| < \epsilon)$$ which proves the first part where $k>0$.

Now if $k<0$, then $\frac{1}{m^k} = m^k > \frac{1}{m}$, so $\left|X_n\right|^k > \left|X_n\right|$, and we can't say that $\left|X_n\right|^k$ also converges to $0$,instead, it diverges to $+\infty$. Is this correct?

A:

The original statement (given) says that for any $\varepsilon>0,\ \exists\ N\in\mathbb{N}$ such that 
$$|X_n|<\varepsilon\ \forall\ n\geq N$$
From there, assuming that $k>0$, for any $\varepsilon^{1/k}>0$ (a different epsilon, of course), we also have that $\exists\ N\in\mathbb{N}$ such that 
$$|X_n|<\varepsilon^{1/k}\ \forall\ n\geq N$$
Noting that for large $n$, $|X_n|<1$, we can legitimately say that 
$$|X_n|^k<\varepsilon\ \forall\ n>N$$
It seems you have a good grip on this. Now, for the second result, let $\ell=-k<0$, so that $\varepsilon^{1/\ell}=\frac{1}{\varepsilon^{1/k}}$. We have, from the given, that for any $\frac{1}{\varepsilon^{1/k}}>0$ and some $N\in\mathbb{N}$,
$$|X_n|<\frac{1}{\varepsilon^{1/k}}\ \forall\ n\geq N$$
Since both sides are positive, this permits the rearrangement
$$\varepsilon^{1/k}<\frac{1}{|X_n|}\ \forall\ n\geq N$$
Again, with both sides positive, we can also say
$$\varepsilon<\frac{1}{|X_n|^k}\ \forall\ n\geq N$$
Considering $\varepsilon$ arbitrarily large this time, we can see that by the quantity diverges as $n\to\infty$. The final step is to note that we can simply rewrite the last step as 
$$\varepsilon<|X_n|^{\ell}\ \forall\ n\geq N$$
where $\ell=-k$.

