Show that $\lim_{x \to c} x^{3}=c^{3}$ Please check my proof 
I will use the fact that
$\lim_{x \to c}x^{3}=c^{3}$ equivalent $\lim_{x \to c}x\lim_{x \to c}x\lim_{x \to c}=(c)(c)(c)$
for $\lim_{x \to c} x=c$ we 
let $\epsilon >0$ and $\delta >0$
we have
$0<|x-c|<\delta \leftrightarrow |x-c|<\sqrt[3]{\epsilon }$
but in this case we want to proove $\lim_{x \to c}x^{3}=c^{3}$
We will have 
$0<|x-c|<\delta \leftrightarrow |x-c|<\sqrt[3]{\epsilon }$
$0<|x-c|<\delta \leftrightarrow |x-c|<\sqrt[3]{\epsilon }$
now $\lim_{x \to c}x^{3}=c^{3}\rightarrow \lim_{x \to c}x\lim_{x \to c}x\lim_{x \to c}x=(c)(c)(c)$
$0<|x-c|<\delta \leftrightarrow |x-c||x-c||x-c|<\sqrt[3]{\epsilon }\sqrt[3]{\epsilon }\sqrt[3]{\epsilon }=\epsilon $
 A: $\textbf{Hint}$: Use difference of cubes. 
$$ x^3 - c^3 = (x-c)(x^2 + xc + c^2)$$
The right term in the product is a polynomial i.e continuous and so it is bounded on $[c- \epsilon, c+ \epsilon]$ say by $M$. It now follows that,
$$ \forall \epsilon>0, |x-c|<\delta := \dfrac{\epsilon}{M} \Rightarrow |x^3 - c^3|< \frac{\epsilon}{M} \cdot M = \epsilon$$
$\textbf{Edit}$: I should not use the fact that $x^2+xc+c^2$ is a polynomial and so continuous as another user points out, since you could just say $x^3$ is continuous and so there is nothing to prove. Instead, get a bound for the polynomial in this manner, suppose $x\in (c-\epsilon, c+ \epsilon)$ then given $\epsilon >0$ you can always choose $M$ such that $M > |c| + \epsilon$ and so,
$$ |x^2 + xc + c^2| < M^2 + M^2+M^2 = 3M^2$$
Hence instead choose,
$$\delta = \frac{\epsilon}{3M^2}$$
A: Here is an argument which does not use continuity. You are on the right track by observing that properties of limits give you
$$\lim_{x\to c}(x^3)=\left(\lim_{x\to c}x\right)^3\;.$$
Notice that its enough to calculate the limit $\lim_{x\to c}x$. We can simply plug that answer into the above equation and it will calculate the limit for us. 
So we can forget about the rest of the problem and try to just calculate the limit $\lim_{x\to c}x$. But this is really easy to do using $\epsilon$, $\delta$ style argument.
Let $\epsilon>0$ be given and choose $\delta=\epsilon$. Then if $\vert x-c \vert<\delta$...well then $\vert x-c\vert<\delta=\epsilon$. Hence, by definition
$$\lim_{x\to c}x=c\;.$$
Now going back to the original problem, we conclude that
$$\lim_{x\to c}(x^3)=\left(\lim_{x\to c}x\right)^3=c^3\;.$$
A: More generally,
to show that
$\lim_{x \to c} x^n
= c^n$
for positive integral $n$.
$x^n-c^n
=(x-c)\sum_{k=0}^{n-1} x^k c^{n-1-k}
$.
Since we are considering
$x$ close to $c$,
we can assume that
$|x-c| < 1$
so that
$|x| < |c|+1$.
Then,
for such $x$,
$\begin{array}\\
|\sum_{k=0}^{n-1} x^k c^{n-1-k}|
&\le \sum_{k=0}^{n-1} |x^k c^{n-1-k}|\\
&\lt \sum_{k=0}^{n-1} |(|c|+1)^k c^{n-1-k}|\\
&\lt n(|c|+1)^{n-1}\\
\end{array}
$
Therefore
$\begin{array}\\
|x^n-c^n|
&=|(x-c)\sum_{k=0}^{n-1} x^k c^{n-1-k}|\\
&=|x-c|\,|\sum_{k=0}^{n-1} x^k c^{n-1-k}|\\
&\lt|x-c|\,|n(|c|+1)^{n-1}|\\
\end{array}
$
so that,
if
$|x-c|
< \min\left(1, \dfrac{\epsilon}{n(|c|+1)^{n-1}}\right)
$,
then
$|x^n-c^n|
\lt \epsilon$.
