Limiting Behaviour of Mean Value Theorem ($\theta \to \frac12$ as $h \to 0$)

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/.

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,

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