Misunderstanding problem solution

[farleyknight]

Hey all,

Homework Statement

This isn't really homework, just looking through a particular book (Spivak's Calculus, 3rd ed) and trying to see if my understanding of a problem is clear or not.

Problem 9, page 107, for those of you following along at home.

Prove that:

\lim_{x \to a} f(x) = \lim_{h \to 0} f(a + h)

Homework Equations

Now, the book gives the following answer:

Let l = \lim_{x \to a} f(x) and define g(h) = f(a + h). Then for every \epsilon > 0 there is a \delta > 0 such that, for all x, if 0 < |x - a| < \delta then |f(x) - l| < \epsilon. Now, if 0 < |h| < \delta then 0 < |(h + a) - a| < \delta so |f(a + h) - l| < \epsilon. This inequality can be written |g(h) - l| < \epsilon. Thus \lim_{h \to 0} g(h) = l, which can be also written \lim_{h \to 0} f(a + h) = l. The same sort of argument shows that if \lim_{h \to 0} f(a + h) = m, then \lim_{x \to a} f(x) = m. So either limit exists if the other does, and in this case, they are equal.

The Attempt at a Solution

Now it appears to me that the author is making the a priori assumption that:

\lim_{x \to a} f(x) = L and \lim_{h \to 0} f(a + h) = L

and using that in his proof to show the two are equal. But I'm doubtful this is a legitimate jump.

Let's do a different setup and I'll show you want I mean.

Given \lim_{x \to a} f(x) = L and \lim_{h \to 0} f(a + h) = M , prove \lim_{x \to a} f(x) = \lim_{h \to 0} f(a + h)

0 < |x - a| < \delta implies |f(x) - L| < \epsilon

Now if

0 < |h| < \delta then

0 < |(h + a) - a| < \delta implies |f(a + h) - M| < \epsilon

Oh wait, we can't go any further.. The two are not necessarily equal because L might not equal M.

So I'm pretty sure the book is right and I'm wrong.. so what's the error in my logic?

- Rob

 

[Dick]

The reason why |f(a+h)-L|<epsilon isn't because of a separate assumption that f(a+h) has a limit. It's because you can substitute h+a for x in the first limit. h+a is one of the values of x such that |x-a|<delta.