Epsilon delta theorem proof Prove Theorem 2.5(iii), namely, that if $f$ is continuous at $a$ and $\lambda$ is a scalar, then $\lambda\cdot f$ is continuous at $a$.

(a) Complete the first principles statement of what we are required to prove, namely, that $λ\cdot f$ is continuous at $a$:
  ‘For every $\varepsilon > 0$, there is $\delta > 0$ such that ... ’.

Okay so I have that 'For every $\varepsilon > 0$, there is $\delta > 0$ such that
$$0 < |x-a|<\delta \Rightarrow |\lambda f(x)-\lambda f(a)|<\varepsilon$$

(b) When $\lambda = 0$, how may $\delta$ be chosen to ensure that (a) holds?

So I have that
$$|\lambda f(x)-\lambda f(a)|<\varepsilon \iff \lambda|f(x)-f(a)|<\varepsilon \iff |f(x)-f(a)|<\varepsilon /\lambda$$
But here $\lambda = 0$, so this is not defined. I'm not sure how this relates to the $\delta$ I have to choose, or how to choose $\delta$.

(c) When $\lambda$ does not equal 0, how may $\delta$ be chosen to ensure that (3a) holds?

I have that
$$|\lambda f(x)-\lambda f(a)|<\varepsilon \iff \lambda|f(x)-f(a)|<\varepsilon \iff |f(x)-f(a)|<\varepsilon /\lambda$$
Does this imply that the $\delta$ I need to choose is $\min(1,\varepsilon/\lambda)$? And how does all of this relate to what I want to prove, that if $f$ is continuous at $a$ and $\lambda$ is a scalar, then $\lambda\cdot f$ is continuous at $a$?
 A: You are on track, but miss a point: We do not need equivalence at some of these steps.
For example, when $\lambda=0$:
We know that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$ and are looking for a delta (called $\tilde\delta$ to distinguish it from the $\delta$ just mentioned) such that $|x-a|<\tilde\delta$ implies  $|\lambda f(x)-\lambda f(a)|<\epsilon$. You are trying to make an equivalence transform: $|0 \cdot f(x)-0\cdot  f(a)|<\epsilon$ is equivalent to (or "if and only if") $|f(x)-f(a)|<\epsilon/0$ (which does not work). But that is not what we need here. All we need is that $|0 \cdot f(x)-0\cdot  f(a)|<\epsilon$ whenever (some condition). Note that $|0 \cdot f(x)-0\cdot  f(a)|$ is always $=0$, so from the given fact that $\epsilon>0$, we see that (some condition) can be anything we want. In other words: We can pick any (positive) value we like for $\delta'$. So why not just let $\tilde\delta=42$?
You make another mistake in the $\lambda\ne 0$ case: Here $|\lambda f(x)-\lambda f(a)|<\epsilon$ is not equivalent to $|\lambda f(x)-\lambda f(a)|<\epsilon/\lambda$  but rather to $|\lambda f(x)-\lambda f(a)|<\epsilon/|\lambda|$. And then still we cannot pick $\delta$ the way we do. Rather, we must start afresh from a different $\epsilon$! Indeed, we let $\epsilon'=\epsilon/\lambda$. From the continuity of $f$ at $a$, we know that for this $\epsilon'>0$ there exists some $\delta'>0$ such that $|x-a|<\delta'$ implies $|f(x)-f(a)|<\epsilon'$. Note that $\delta'$ may have little to do with the $\delta$ we had for $\epsilon$. But at least we can pick this $\delta'$ as our $\tilde \delta$.
A: Ok for $(a)$.
(b) If $\lambda=0$, you can actually choose any positive $\delta$, because $|\lambda f(x)-\lambda f(a)|$ will always hold, for any $x$, as $\lambda f=0$.
(c) You have to use the fact that $f$ is continuous at point $a$, which implies that
$$\forall \epsilon>0, \exists \delta>0,\, |x-a|\lt \delta\implies |f(x)-f(a)|\lt \frac{\epsilon}{|\lambda|}$$ 
So you can choose this particular $\delta$ (however be aware that it still depends on $\epsilon$ and on $a$).
