# Thread: Commutativity of limit using epsilon-delta

1. ## Commutativity of limit using epsilon-delta

I'm trying to write a delta-epsilon proof that:

$\lim_{x\to a}kf(x)=k\left(\lim_{x \to a}f(x)\right)=b\quad\text{(k is constant)}$

I know that $\lim_{x\to a}kf(x)$ exists if $\forall\epsilon>0\exists\delta>0$ such that:

$0<|x-a|<\delta \Rightarrow |kf(x)-b|<\epsilon$

And $k\left(\lim_{x\to a}f(x)\right)$ exists if $\forall\epsilon>0\exists\delta>0$ such that:

$0<|x-a|<\delta \Rightarrow k|f(x)-b|<\epsilon$

I don't understand how I can write a proof using an arbitrary function and limit. So far I have only been proving specific limits with the epsilon delta method, not writing any algebraic proofs like this. Can someone offer some help? Thanks

Edit: I had a thought, but I'm not sure I wrote that second one out right, cause wouldn't I simply have to prove: $|kf(x)-b|=k|f(x)-b|$ (which isn't true)?

2. We will assume $k\not = 0$.

Since $\lim_{x\to a}f(x) = b$ it means for any $\epsilon > 0$ we can find $\delta > 0$ so that $0 < |x-a| < \delta \implies |f(x) - b| < \epsilon$.

We want to show $\lim_{x\to a} kf(x) = kb$. Let $\epsilon > 0$ but then $\tfrac{\epsilon}{|k|} > 0$. This means by above there is $\delta > 0$ so that $0 < |x-a| < \delta \implies | f(x) - b | < \tfrac{\epsilon}{|k|}$. Multiply by $|k|$ to get $|kf(x) - bk| < \epsilon$ whenever $0 < |x-a| < \delta$.

3. Thanks alot, I understand it now.