# Prove $\lim_{n \to \infty}ka_n = k\lim_{n \to \infty} a_n$

So I am given this to prove: $$\lim_{n \to \infty}ka_n = k\lim_{n \to \infty} a_n$$

This problem seems trivial to solve, but bear with me here. Would I be able to solve it by using the definition of a limit of a sequence? If not that, then how would I be able to prove it? I am really stuck on getting it started off.

Any help is greatly appreciated.

Let $\epsilon > 0$ be given. By definition, we can choose some $N$ so that
$$| a_n - L | < \epsilon /k , \; \; \; for \; n > N$$
You want to show $k a_n \to k L$.
But, $| ka_n - k L | = k |a_n - L| < \epsilon$
Suppose that $\lim_{n\to \infty} a_n=S$. Now our goal is to prove that $$\lim_{n\to \infty} ka_n=kS.$$
For any $\epsilon >0$, since $\lim_{n\to \infty} a_n=S$, then there exists $N>0$ such that when $n>N$, $|a_n-S|<\frac{\epsilon}{|k|}$. So $$|ka_n-kS|=|k||a_n-S|<|k|\epsilon=\epsilon.$$