$\lim_{n \to\infty} p_n = p$ implies $\lim_{n \to \infty}p_n^3 = p^3$ In an example from my lecture notes, I have that if $( p_n )_n$ is a sequence and $\displaystyle\lim_{n \to \infty} p_n = p$, then $\displaystyle\lim_{n \to \infty} p_n^3 = p^3$.
I don't understand all of this example, though I do understand it when $p=0$. 
Now let $p\neq0$. Then there exists $n_1\in \mathbb{N}$ such that $|p_n-p|=|p|/2$ for all $n\geq n_1$. This is also a theorem in our book. I don't understand it. Why is this true and what is its purpose?
From that we have this line of reasoning: for all $n \geq n_1$
$$\left| |p_n|-|p| \right|\leq|p_n-p|=\tfrac{|p|}{2}\longrightarrow \tfrac{|p|}{2}<|p_n|<\tfrac{3|p|}{2}.$$ 
I don't fully understand this.
This follows: (I didn't understand it but after typing it I get it!) Thought I'd leave it in.
$$\begin{align}
|p_n^3-p^3| 
&= |p_n-p| \cdot |p_n^2+p_n\cdot p+p^2| \\\
& \leq |p_n-p| \cdot (|p_n^2|+|p_n| \cdot |p|+|p^2|) \\
&\leq |p_n-p| \cdot (\tfrac{9|p|^2}{4}+\tfrac{3|p|^2}{2}+|p^2|) \\
&= \tfrac{19|p|^2|p_n-p|}{4}.
\end{align}$$
Then $n_3$ and $\epsilon$ is defined and the proof is wrapped up. Can anyone please explain the parts above I don't understand. I'd really appreciate it! Thank you!
 A: The statement $\displaystyle\lim_{n \to \infty}p_n = p$ means that "as $n$ becomes large, $|p_n-p|$ becomes arbitrarily small". 
We wish to prove $\displaystyle\lim_{n \to \infty}p_n^3 = p^3$, i.e. "as $n$ becomes large, $|p_n^3-p^3|$ becomes arbitrarily small". 
Now, note that $|p_n^3-p^3| = |(p_n-p)(p_n^2+p_np+p^2)| = |p_n-p| \cdot |p_n^2+p_np+p^2|$. 
As $n$ gets arbitrarily large, $|p_n-p|$ becomes arbitrarily small. So, if we can show that $|p_n^2+p_np+p^2|$ doesn't become arbitrarily large (or do other weird stuff), then the product $|p_n-p| \cdot |p_n^2+p_np+p^2| = |p_n^3-p^3|$ will become arbitrarily small, which is what we want. 
Now more formally, since $\displaystyle\lim_{n \to \infty}p_n = p$ the definition of the limit says that for any $\epsilon > 0$, there exists an $n_1$ such that $\forall n \ge n_1$, $|p_n-p| < \epsilon$. Since this is true for all $\epsilon > 0$, it must be true for $\epsilon = |p|/2$, that is there exists an $n_1$ such that $\forall n \ge n_1$, we have $|p_n-p| < |p|/2$. If $p > 0$, then this becomes $-p/2 < p_n-p < p/2$, i.e. $p/2 < p_n < 3p/2$. If $p < 0$, then this becomes $p/2 < p_n-p < -p/2$, i.e. $3p/2 < p_n < p/2$. In either case $|p_n| < 3|p|/2$. 
Using this inequality along with the triangle inequality, we get the following:
$|p_n^2+p_np+p^2| \le |p_n^2|+|p_np|+|p^2| < \frac{9}{4}|p|^2+\frac{3}{2}|p|^2+|p|^2 = \frac{19}{4}|p|^2$. 
Therefore, $|p_n^3-p^3| = |p_n-p| \cdot |p_n^2+p_np+p^2| < \frac{19}{4}|p|^2|p_n-p|$. 
Now, intuitively, as $n$ gets arbitrarily large, $|p_n-p|$ gets arbitrarily small, and $\frac{19}{4}|p|^2$ is just a constant. Hence $\frac{19}{4}|p|^2|p_n-p|$ gets arbitrarily small. Then since $|p_n^3-p^3| < \frac{19}{4}|p|^2|p_n-p|$, we know that $|p_n^3-p^3|$ also gets arbitrarily small. To formalize this last part, you just need to apply the definition of the limit. 
A: Explaining what Alan meant: if $\lim_{x}f(x) =a <\infty$ and $ \lim_{x}g(x) = b <\infty$ and both $f$ and $g$ are continuous functions, you can write $\lim_{x}fg = \lim_{x} f \lim_{x} g = ab$, so in your case just write the limit as three products and you get the result.  
