I'm currently taking an introduction to Calculus course and I've come across the following identity:
How would one come up with this? My best guess is using L'Hospital's Rule on $$\lim_{x\rightarrow a}{\frac{f(x)-f(a)}{x-a}}$$
but I'm not very sure how, since differentiating both the numerator and denominator merely yields
$$\lim_{x\rightarrow a}{f'(x)} = f'(a)$$
Answer
The result holds under the weaker assumption that $f''(a) $ exists (other answers assume the continuity of $f''$ at $a$ or even more). Also note that under this weaker assumption it is not possible to apply L'Hospital's Rule on the expression under limit in question and hence a slight modification is required.
By definition of derivative we have $$\lim_{x\to a} \frac{f'(x) - f'(a)} {x-a} =f''(a)\tag{1}$$ Adding this to the limit in question it is clear that our job is done if we can establish that $$\lim_{x\to a} \frac{f(x) - f(a) - (x-a) f'(a)} {(x-a)^2}=\frac{f''(a)}{2}\tag{2}$$ And the above limit is easily evaluated by a single application of L'Hospital's Rule. Applying it on the fraction on left side we get a new fraction $$\frac{f'(x) - f'(a)} {2(x-a)}$$ which clearly tends to $f''(a) /2$ (via $(1)$) and hence the fraction on left side of $(2)$ also tends to the same value and the identity $(2)$ is established.
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