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This was posted in another question and below is the link.

Prove Theorem 2.5(iii), namely, that if $f$ is continuous at $a$ and $λ$ is a scalar, then $λ⋅f$ is continuous at $a$ i.e. $$0<|x−a|<δ⇒|λf(x)−λf(a)|<\varepsilon$$

Here $|λf(x)−λf(a)|<\varepsilon$ is not equivalent to $|λf(x)−λf(a)|<\frac{\varepsilon}{λ}$ but rather to $|λf(x)−λf(a)|<\frac{\varepsilon}{|λ|}$

My question is regarding the last inequality, shouldn't this be $|λf(x)−λf(a)|<\varepsilon$ instead?

Epsilon delta theorem proof

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stackdsewew
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I think the last part:

Here |λf(x)−λf(a)|<ϵ is not equivalent to |λf(x)−λf(a)|<ϵ/λ but rather to |λf(x)−λf(a)|<ϵ/|λ|

should read

Here |λf(x)−λf(a)|<ϵ is not equivalent to |f(x)−f(a)|<ϵ/λ but rather to |f(x)−f(a)|<ϵ/|λ|.

It is just saying that when you divide each side of the equation thorough by λ on each side, you can only do so by taking the modulus, |λ|. Otherwise if λ<0 the inequality would give a modulus (strictly positive) value on the LHS and a negative on the RHS which would clearly be wrong.

user40130
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  • The modulus part I understood, but I want to know why there is still a lambda on the LHS. Because we divided epsilon by lambda, shouldn't the lambda on the LHS disappear. – stackdsewew May 10 '16 at 15:38
  • Oh it seems you picked up on that too. Cheers – stackdsewew May 10 '16 at 16:00
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