Proving a limit with a constant Suppose that $x_n \rightarrow a$ and $c \in \mathbb{R}$. Prove that $cx_n \rightarrow ca$.
I'm comfortable doing so using the multiplicative properties of limits and utilizing the limit of a constanst, but how can I do this using the definition of limits?
Thanks in advance.
 A: HINT: There’s certainly no problem if $c=0$. If $c\ne 0$, note that $|ca-cx_n|=|c||a-x_n|$. If you make $|a-x_n|<\frac{\epsilon}{|c|}$, you automatically make $|ca-cx_n|<\epsilon$.
A: $$\forall\,\epsilon >0\;:\;\;\;|cx_n-ca|=|c||x_n-a|<\epsilon\Longleftrightarrow |x_n-a|<\frac{\epsilon}{|c|}\,\,,\,c\neq 0$$
But you can find 
$$N\in\Bbb N\,\,\,s.t.\,\,\,n>N\Longrightarrow |x_n-a|<\frac{\epsilon}{|c|}\ldots$$
The case $\,c=0\,$ I leave to you.
A: Assume that $x_n \to a$ as $n\to \infty$. Assume $c\neq 0$. You want to prove that for all $\epsilon>0$ there is a $n_0$ such that for $n\geq n_0$ you have $\lvert cx_n - ca\rvert < \epsilon$.
So we start with:
Let $\epsilon >0$ be given. Then note that
$$
\lvert cx_n - ca\rvert = \lvert c\rvert\cdot\lvert x_n - a\rvert.
$$
Now since $x_n \to a$ (for $n \to \infty$) you know by definition that there is an $n_0$ such that for $n\geq n_0$ you have 
$$
\lvert x_n - a\rvert < \frac{\epsilon}{\lvert c \rvert}
$$
But this last inequality is exactly what you want because you can multiply by $\lvert c\rvert$ on both sides and there by get what you need.
