Spivak proof for limits, rigorous, difficult.

Question

Prove that $\displaystyle \lim_{x\to a} f(x) = L \space \text{if and only if} \space \lim_{x\to a} [f(x) - L] = 0$ Provide a rigorous proof.

I am not sure what he has given to us.

Is $\displaystyle \lim_{x\to a} f(x) = L$ true?

How would we do this?

Answer 1

Nothing has been given (aside from the usual implicit things like the axioms of complete ordered fields): you're being asked to prove $P \leftrightarrow Q$.

Note that if you were asked to prove $P \to Q$, then nothing has been given either. However, a common proof technique is to show that, if you were given $P$, then you could conclude $Q$.

More precisely, let me write $P \vdash Q$ to means "you can prove $Q$ by assuming $P$". There's a theorem of logic that says $\vdash P \to Q$ (i.e. we can prove $P \to Q$ without being given anything) if and only if $P \vdash Q$.

Thus we have a proof technique (a "direct" proof) for proving implications:

• Do the following:
  • Assume $P$
  • ...
  • Conclude $Q$
• Conclude $P \to Q$

By the above notation, I mean that we're only assuming $P$ throughout the part of the argument covered by the first top-level bullet point. By the time we reach the last top-level bullet point, we are no longer assuming $P$.

Clearly, the difference between $P \to Q$ and $P \vdash Q$ is subtle, and the distinction is not often made in common mathematical parlance (and frequently, it's not worth making the distinction).

One of the usual proof techniques for proving $P \leftrightarrow Q$ is:

• Do the following
  • ...
  • Conclude $P \to Q$
• Do the following
  • ...
  • Conclude $Q \to P$
• Conclude $P \leftrightarrow Q$

although you could reorganize it as

• Do the following
  • Assume $P$
  • ...
  • Conclude $Q$
• Do the following
  • Assume $Q$
  • ...
  • Conclude $P$
• Conclude $P \leftrightarrow Q$

(as before, note that we are only assuming $P$ while working through the first top-level bullet point: we drop that assumption when we move onto the second and third top-level bullet points)

The first bullet point is often described as "proving the forward direction" or "proving $\to$" or even just prefixed with "$(\to)$" to say what you're trying to do. When working through that point (but nowhere else), it wouldn't be too misleading to call $P$ a given.

The second is "proving the reverse/converse/backwards direction" or $(\leftarrow)$ or somesuch.

Answer 2

Let g(x) = f(x) - L. First assume lim x→a f(x)=L. Then for all ε > 0, there is a δ > 0 such that whenever 0 < |x - a| < δ, we have |f(x) - L| = |g(x) - 0| < ε. Hence lim x→a g(x) = 0, or lim x→a [f(x) - L] = 0. Second assume lim x→a [f(x) - L] = 0. Then for all ε > 0, there is a δ > 0 such that whenever 0 < |x - a| < δ, |f(x) - L - 0| = |f(x) - L| < ε. This is the criterion for lim x→a f(x)=L. The foregoing establishes that lim x→a f(x)=L⇔lim x→a [f(x) - L] = 0.