# If $f$ is continous at $c$, prove $\lim_{h \to 0} (\inf \,\{f(x)\mid c \leqslant x \leqslant c+h\})=f(c)$(duplicated)

Let $$f$$ be continous at $$c$$. Prove $$\lim_{h \to 0} \left(\inf \,\{f(x)\mid c \leqslant x \leqslant c+h\}\right)=f(c)$$

This fact is used in Spivak's book to prove 1nd Fundamental Calculus Theorem.

This question is duplicated, I have done it previously but i didn't understand yet

The inf is bounded from below for every $$h > 0$$ because f is continous (can you show why?). Assuming $$\lim_{h \rightarrow \infty} \inf \{ f(x) : c \leq x \leq c + h \} =: K \neq f(c),$$ then there exits some sequence $$(x_n)_n$$ such that for all $$n \in \mathbb{N} : x_n > c, \lim_{n \rightarrow \infty} x_n = c$$ and $$\lim_{n \rightarrow \infty} f(x_n) = K \neq f(c) = f(\lim_{n \rightarrow \infty} x_n)$$ a contradiction to the continuity of $$f$$.
Actually, that's not the only thing Spivak uses in his proof. He first considers the case $$h > 0$$, then $$h < 0$$, and then claims that $$\lim \limits_{h \to 0} m_h = 0$$. So, to take into account both signs of $$h$$, lets do the following: suppose $$a < c < b$$, and for any $$h \in \Bbb{R}$$ , define the set $$$$A_h = \{f(x): a \leq x \leq b \quad \text{and} \quad |x-c| \leq |h| \}.$$$$ First, since $$f$$ is integrable, it is by definition bounded, so for all $$h \in \Bbb{R}$$, $$A_h$$ is a non-empty, bounded set hence $$m_h := \inf A_h$$ exists. We now wish to use continuity of $$f$$ at $$c$$ to prove $$\lim \limits_{h \to 0}m_h = f(c)$$.
To do this, let $$\varepsilon > 0$$ be arbitrary. By assumption, $$f$$ is continuous at $$c$$, so there is a $$\delta > 0$$ such that for all $$x \in [a,b]$$, if $$|x-c| < \delta$$, then $$|f(x) - f(c)| < \varepsilon$$. Now, let $$0 < |h| < \delta$$ be arbitrary. Then, for every $$x \in [a,b]$$ which satisfies $$|x-c| \leq |h|$$, we have that (since $$|h| < \delta$$) $$$$f(c) - \varepsilon < f(x) < f(c) + \varepsilon.$$$$ What this says is that $$f(c) - \varepsilon$$ is a lower bound for $$A_h$$ and $$f(c) + \varepsilon$$ is an upper bound for $$A_h$$. Hence, it follows by definition of infimium that $$$$f(c) - \varepsilon \leq m_h \leq f(c) + \varepsilon$$$$ Or equivalently, $$|m_h - f(c)| \leq \varepsilon$$. This completes the proof, because we have shown that given any $$\varepsilon > 0$$, there is a $$\delta > 0$$ such that for all $$h \in \mathbb{R}$$, if $$0 < |h| < \delta$$, then $$|m_h - f(c)|\leq \varepsilon$$.
Note: From my proof, we can only conclude that $$|m_h - f(c)| \leq \varepsilon$$, not $$|m_h - f(c)| < \varepsilon$$. But this still shows the desired limit because this is true for every $$\varepsilon > 0$$. If this isn't clear to you, see one of the exercises in Chapter 5 of Spivak (it's probably 25 or 26).
Next, for $$c=a$$ or $$c=b$$ you only have to modify the proof slightly; and for the proof of $$\lim \limits_{h \to 0} M_h = f(c)$$, you just have to replace $$m_h$$ with $$M_h$$ everywhere above, and the same proof works.