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I have a function, $f$, which is continuous on $[a, a+h]$ and differentiable on $(a, a+h)$. By the mean value theorem, there exists a $\theta \in (0,1)$ such that, $$f(a+h) - f(a) = h f'(a+\theta h)$$

Clearly, $\theta$ generally depends on $h$. The goal is to prove that, given that $f''(a)$ exists and is non-zero, $$\lim_{h \to 0} \theta = 1/2$$

Note: Using Taylor's theorem this is not too difficult, however, as per the source of the question, there is a solution that uses nothing more than the meant value theorem (including Cauchy's), l'Hopital's rule, and whatever else would generally be considered more 'basic' (e.g. definition of derivative, rules of limits, etc.).

Edit: If the question is not too clear, an example of a specific case of what I seek to prove can be found here (the very last part of the post): http://www.stumblingrobot.com/2015/09/27/prove-an-alternate-expression-for-the-mean-value-formula/.

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  • $\begingroup$ Hint: divide by $h^2$ $\endgroup$ – TetstName123 Mar 7 '17 at 19:45
  • $\begingroup$ @M.Winter Nope, definitely not - this is true for the general function $f$, as described in the question. As nice as it would be, unfortunately, I cannot just 'solve' the question by essentially changing it to a much easier, less general one. Also, this is essentially the same idea as what is already in the link at the end of the question... $\endgroup$ – John Don Mar 7 '17 at 19:49
  • $\begingroup$ Express everything in terms of $f'(x)$ instead of $f (x)$. For instance $f$ becomes the primitive of $f '$ (if you can use primitives/integrals). In this way it should become easier to visualize. $\endgroup$ – Del Mar 7 '17 at 22:43
  • $\begingroup$ @Del I completely agree - the key point is about being able to visualise things more easily. The solution, now that I know it, is not hard at all (perhaps even 'obvious') - I was just stuck in a bit of a rut as I think I had stared at the problem for too long. However, forcing things into a format that you know will get you closer to the end goal (i.e. writing things in terms of $f'(x)$) helps overcome this as it makes it much easier to spot what the next 'obvious' step should be as it is easier to get things into a more immediately recognisable form. $\endgroup$ – John Don Mar 7 '17 at 23:25
  • $\begingroup$ @TetstName123 Thanks for the suggestion - while it wasn't obvious how to proceed from you suggestion at first, the other crucial 'elusive' step turned out to be subtracting $f'(a)$ from each side first (kind of obvious in retrospect), and then everything falls into place pretty routinely (see RRL's answer, below). $\endgroup$ – John Don Mar 7 '17 at 23:36
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Note that

$$\frac{f(a+h) - f(a) - h f'(a)}{h^2} = \frac{(f'(a+ \theta h)-f'(a))h}{h^2} \\ = \frac{f'(a+ \theta h) - f'(a)}{\theta h}\theta.$$

Taking the limit of both sides using L'Hospitals rule on the left we get

$$ \lim_{h \to 0} \frac{f'(a+h) - f'(a)}{2h}= \frac{1}{2}f''(a) = f''(a) \lim_{h \to 0} \theta.$$

Thus,

$$\lim_{h \to 0} \theta = \frac{1}{2}.$$

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  • $\begingroup$ Thank you so much for this answer - I was stupidly close to this elusive answer for so long (I think I had been staring at it for too long) - just a very minor point (I suppose it comes down to style/personal preference): would it not be just a tad clearer to re-write $\int_a^{a+h}f'(x) \, dx$ just simply as $f(a+h) - f(a)$?... Feel free to ignore this if you disagree, but thanks again! $\endgroup$ – John Don Mar 7 '17 at 23:20
  • $\begingroup$ Sure - either way. I just started down that path as I was trying to figure this out (saw comment suggesting integral) and did not get around to changing it. It should be clear why $f''(a) = \lim_{h \to 0} \frac{f(a+\theta h ) - f(a)}{\theta h} = f''(a)$ since $\theta$ is at least bounded. $\endgroup$ – RRL Mar 7 '17 at 23:25
  • $\begingroup$ Yes, I see what you mean - in fact, I think it shows a bit more clearly your motivation for how you reached this solution. And yes, I agree that last limit is of course obvious - for example, since $\theta$ is bounded, we may replace '$\lim_{h \to 0}$' by '$\lim_{\theta h \to 0}$' (by the usual rules of limits), and then we have the usual definition of the derivative. $\endgroup$ – John Don Mar 7 '17 at 23:30
  • $\begingroup$ Please do - I'm always interested in nice problems. $\endgroup$ – John Don Mar 7 '17 at 23:51
  • $\begingroup$ @JohnDon: A natural follow-up question is what if $f''(a) = 0$ or does not exist. Here is a similar question, but for the integral MVT. $\endgroup$ – RRL Mar 8 '17 at 2:14

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