Example: Let $f$ be continuous at $c$, and let $m_h = \inf \{ f(x) : c \le x \le c + h \}, M_h = \sup \{ f(x) : c \le x \le c + h \}$. Then $\displaystyle\lim_{h\to0} m_h = \displaystyle\lim_{h\to0} M_h = f(c)$, but what is the formal justification for this?

Really, what we have is an interval $I = [c,c+h]$ and then $\lim_{h\to0} I = c$, correct? But still, this seems non-rigorous in the sense that the $\varepsilon-\delta$ definition of a limit existing that I know of won't support taking the limit of an interval in that form. Of course, intuitively it makes sense.

But also, we don't necessarily require a closed interval. For example: The MVT states that if $f$ is differentiable on $[a,b]$, then $\exists c \in (a,b)$ such that $f(b) - f(a) = f'(c)(b - a)$. Now when solving a certain problem in Spivak's Calculus, he used the fact that if $f$ is differentiable on $[a,b]$, then for $x \le b$, $f(x) - f(a) = f'(\zeta)(x - a)$ for some $\zeta \in (a,x)$. Then he argued that as $x \to a$, $f'(\zeta) \to f'(a)$. But how does this make sense? $\zeta \in (a,x)$, so as $x \to a$, we get the interval approaching $(a,a)$.

  • $\begingroup$ You don't need "limits of intervals" to answer this question, and you don't need to go into the MVT to understand it. A straightforward proof using the definitions of continuity, of limits, and of infimums is certainly possible. Did you make any attempts in that direction? $\endgroup$
    – Lee Mosher
    Jul 14 at 21:43
  • $\begingroup$ @LeeMosher Yes, in the first example would want to show that $\forall \varepsilon > 0, \exists \delta > 0, 0 < |h| < \delta \implies |\inf \{ f(x) : c \le x \le c + h\} - f(c)| < \varepsilon$. If we let $\alpha$ be the infimum, then $\alpha \le f(c)$ by definition, so we want to show that $f(c) - \alpha < \varepsilon$. The point of my second example is that the open interval goes to $(a,a)$ and so MVT is not even satisfied! $\endgroup$ Jul 14 at 21:46

1 Answer 1


I assume that $f : \mathbb{R} \to \mathbb{R}$ and $h \ge 0$.

Note that $m_h \in \mathbb{R}$. So actually, we could write $g : \mathbb{R}^{\ge 0} \to \mathbb{R}$,

$$g(h) = \text{inf} \{f(x):c \le x \le c+h\}$$

There are no limits of intervals (although you can imagine that $g$ is related to a family of descending intervals). The limit can equivalently be written as $\lim_{h \to 0} g(h)$, where as noted, $g$ is a standard function $\mathbb{R} \to \mathbb{R}$. Similarly for $M_h$.

Hopefully, this answers your main question. With Spivak's proof, the issue is that:

$$x \to a, \; \zeta \in (a,x) \not \Rightarrow \zeta \in \lim_{x \to a} (a,x)$$

even if the limit on the RHS is/was well-defined. Again, we are only concerned with the limit of a function $\mathbb{R} \to \mathbb{R}$, which in this case is $f'$. We seek to show that $\forall \epsilon > 0, \exists \delta >0 $ such that:

$$|x - a| < \delta \implies |f'(\zeta)-f'(a)| < \varepsilon$$

(Or you could use an equivalent formulation in terms of sequences.)

In this case, there is a simpler way to do it:


$$\implies f'(\zeta) = \frac{f(x)-f(a)}{x-a}$$

$$\implies \lim_{x \to a} f'(\zeta) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$

$$ \implies \lim_{x \to a} f'(\zeta) = f'(a)$$

  • $\begingroup$ But why doesn't it matter that $\zeta \in (a,x)\to(a,a)$ as $x \to a$? $\endgroup$ Jul 15 at 11:08
  • $\begingroup$ @politeproofs It doesn't matter because it is not what Spivak claims in the proof. He writes $f'(\zeta) \to f(a)$ as $x \to a$, which is a statement about $f'$, not the interval $(a,x)$. Remember also that we can define limits $x \to a$ of a function $g$ even if $g(a)$ is not defined. Limits are about the behaviour near a certain point, not at the point. $\endgroup$ Jul 15 at 11:25
  • $\begingroup$ But to even get the existence of $f'(\zeta)$, we use the MVT which tells us that $\zeta \in (a,x)$. $\endgroup$ Jul 15 at 11:52
  • $\begingroup$ And another thing: you're right we could write $m_h$ as $g(h)$. But isn't $\lim_{h\to0^+} m_h$ undefined? We need $\lim_{h\to0^-} m_h$ $\endgroup$ Jul 15 at 11:53
  • $\begingroup$ @politeproofs We are only concerned with the limit of $f'(\zeta)$. This is perfectly well-defined even if $ \zeta \in (a,x)$. If you disagree, please explain why the reasoning I gave fails. The issue is just that you are misinterpreting the limit. As for $m_h$, note that if $h < 0$, we are taking the infimum of an empty set. This is unlikely to be what is intended, which is why I assumed at the start of my answer that $h \ge 0$. I do not see any reason why the limit $h \to 0^+$ would be undefined. $\endgroup$ Jul 15 at 16:59

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