Prove $\lim _{ n\rightarrow \infty }{ { x }_{ n }^{ k } } ={ \left( \lim _{ n\rightarrow \infty }{ { x }_{ n } } \right) }^{ k }$ I'm trying to prove that the limit of the sequence $x_n^k$ is the same as the limit of $x_n$ all raised to the $k$th power.  Prove $$\lim _{ n\rightarrow \infty  }{ { x }_{ n }^{ k } } ={ \left( \lim _{ n\rightarrow \infty  }{ { x }_{ n } }  \right)  }^{ k }$$
Assume $\displaystyle\lim_{n\to\infty}x_n=L$ and $\displaystyle\lim_{n\to\infty}y_n=M$. Then, $(x_n)$ is bounded and thus there is a constant $c&gt;0$ such that
$$|x_n|\leq c,\quad\forall\ n\in\mathbb N.$$

Furthermore, given $\varepsilon&gt;0$ there is $n_0\in\mathbb  N$ such that
$$n&gt;n_0\quad\Rightarrow\quad |x_n-L|&lt;\frac{\varepsilon}{2|M|},\quad|y_n-M|&lt;\frac{\varepsilon}{2c} .$$

As a consequence,
$$n&gt;n_0\quad\Rightarrow\quad |x_ny_n-LM|\leq |x_n||y_n-M|+|M||x_n-L|&lt;c\frac{\varepsilon}{2c}+|M|\frac{\varepsilon}{2|M|}=\varepsilon.$$

This proves that
$$\lim_{n\to\infty} (x_n\cdot y_n)=LM=\left(\lim_{n\to\infty} x_n\right)\left(\lim_{n\to\infty} y_n\right)$$
Therefore,
$$\lim_{n\to\infty}x_n^k=\lim_{n\to\infty}(x_n\cdot x_n\cdots x_n)=\left(\lim_{n\to\infty} x_n\right)\left(\lim_{n\to\infty} x_n\right)\cdots \left(\lim_{n\to\infty} x_n\right)=\left(\lim_{n\to\infty} x_n\right)^k$$
For $\epsilon &gt; 0$, there is a $M_1 \in \mathbb{N}$ that satisfies:

$$|x_n - L| &lt; \epsilon, \forall n &gt; M_1$$

So $x_n$ converges to $L$. We need a similar result for $x_n^k$ and $L^k$. So, for any $\epsilon' &gt; 0$, we want to show existence of a $M_2 \in \mathbb{N}$ that satisfies:

$$|x_n^k - L^k| &lt; \epsilon', \forall n &gt; M_2$$

But:

$$|x_n^k - L^k| = |(x_n - L)(x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1})|$$

$$= |x_n - L| |x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|$$

Take $\epsilon = \frac{\epsilon'}{|x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|}$. By hypothesis, there existis $M_1 \in \mathbb{N}$ that satisfies:

$$|x_n - L| &lt; \epsilon, \forall n &gt; M_1$$

Therefore:

$$|x_n^k - L^k| = |x_n - L| |x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|\\ &lt; \frac{\epsilon'}{|x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|} \cdot |x_n^{k-1} + x_n^{k-2}L + \cdots + x_nL^{k-2} + L^{k-1}|$$

$$|x_n^k - L^k| &lt; \epsilon'$$

for all $n &gt; M_1$. Taking $M_2 = M_1$ gives the result.