Help in finding the fallacy in the following proof.

Question

I'm currently studying Michael Spivak's Calculus and stumbled across a problem. I seemed to have proven a ridiculous statement:

Let $f$ be continuous over $\mathbb{R}$. Then given sufficiently small, nonzero $h$, we have $$\lim_{x \to a+h} f(x) = \lim_{x \to a} f(x)$$ for all $a$. This assertion is ridiculous, because it follows from this statement that $f(a+h) = f(a)$ for all $a$. However, I cannot find any error in the following proof:

Suppose $\lim_{x\to a} f(x) = f(a)$. Then for all $\varepsilon > 0$, there exists $\delta > 0$ such that if $|x-a| < \delta$, we have $|f(x) - f(a)| < \varepsilon$. Let $h \ne 0$ be such that $|h| < \delta$. Then, if $|x-(a+h)| < \delta -|h|$, we have $$a+h - (\delta - |h|) < x < a+h + \delta - |h|$$ $$\implies a+h - \delta + |h| < x < a+h + \delta - |h|.$$ $h + |h| \geq 0$ and $h-|h| \leq 0$, so the above chain of inequalities imply $$a - \delta < x < a+ \delta$$ $$\implies |x-a| < \delta$$ and thus $|f(x) - f(a)| < \varepsilon$, which implies $$\lim_{x\to a+h} f(x) = f(a).\blacksquare$$

What is the fallacy in the given proof?

Answer 1

You have the following proposition:

Let $f$ be continuous over $\mathbb{R}$. Then given sufficiently small, nonzero $h$, we have $$\lim_{x \to a+h} f(x) = \lim_{x \to a} f(x)$$ for all $a$.

What does that mean? The only way I can see to make sense of it is to say that it is possible to pick a value of $h$ such that $h > 0$ (though $h$ may be very small), and then, using that same value of $h$ all the time, show that $\lim_{x \to a+h} f(x) = \lim_{x \to a} f(x)$ for all $a.$

The flaw in the proof then is that it does not use the same value of $h$ all the time. For each new value of $a,$ for each possible $\epsilon > 0,$ it finds a $\delta > 0$ and then an $h$ such that $\lvert h\rvert < \delta.$ Each time you try a new $a,$ and each time you try a new $\epsilon$ for this $a,$ you may end up with a different $h.$ In fact, for must functions you will be forced to choose different values of $h$ depending on which value $\epsilon$ takes on.

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If you change the limit to $$\lim_{h \to 0}\left(\lim_{x \to a+h} f(x)\right) = \lim_{x \to a} f(x),$$ where $f$ is any continuous function from $\mathbb R$ to $\mathbb R$ and $a$ is any real number, however, you get a true proposition.

You might be able to argue that by taking the limit as $h \to 0,$ you always have a value of $h$ less than any value of $\delta$ you might need to use. On the other hand, considering that $\lim_{x \to a} f(x) = f(a)$ (because $f$ is continuous) and $\lim_{x \to a+h} f(x) = f(a+h),$ I think the proof can be made easier.