# Misunderstanding problem solution

1. Dec 27, 2008

### farleyknight

Hey all,

1. The problem statement, all variables and given/known data

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:

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

2. Relevant 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.

3. 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

2. Dec 27, 2008

### 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.