Is it true that if $f$ is continuous, $\lim_{h \rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt = f(x)$? [duplicate]

And if so, what theorems/results leads to this conclusion. I have been searching like crazy but I don't know what to look for, I have a vague memory of this result and I am roughly 97% that it is correct.

Solution 1:

If $f$ is continuous, the function $F(x) = \int_a^x f(t)dt$ is differentiable with derivative $F'(x) = f(x)$. Notice that $$\lim_{h\to 0}\frac{\int_x^{x+h}f(t)dt}{h} = \lim_{h\to 0}\frac{\int_a^{x+h}f(t)dt - \int_a^x f(t)dt}{h} = \lim_{h\to 0}\frac{F(x+h) - F(x)}{h}$$ and use the definition of the derivative.

It has been pointed out by Stephen Donovan that perhaps this problem arose in proving the fundamental theorem of calculus. In that case, the comments on the original post detail some methods of solution.

Solution 2:

There is no need to invke the FTC for this. Note that, by the monotonicity of the integral, $$h\min_{x\le s\le x+h}f(s) = \int_x^{x+h} \min_{x\le s\le x+h}f(s)\,dt \le \int_x^{x+h}f(t)\,dt$$ Similarly, $$h\max_{x\le s\le x+h}f(s) = \int_x^{x+h}\max_{x\le s\le x+h}f(s)\,dt \ge \int_x^{x+h}f(t)\,dt$$ Now divide these inequalities by $h$ and let $h\rightarrow 0$. By continuity, $$\lim_{h\rightarrow 0} \min_{x\le s\le x+h}f(s) = f(x)$$ with a similar result for the $\max$. Combinining the resulting inequalities yields the result.