$$\lim_{x \to 0}\frac{f(x)-f(0)}{x^2} = \frac{f''(0)}{2}\quad (f'(0) = 0)$$ It seems quite a rudimentary problem, but I can't find an appropriate solution without using L'hospital's rule and Maclaurin series. Is it possible that a problem can not be solved without them?
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What do you mean by "is it possibile that a problem can not be solved without them?". In the worst case you just use the same hypotheses of those theorems (basically the MVT and differentiability up to a certain order) to get to the same result without any mention to the theorems. (Btw, I did not downvote, I actually think this is an important point.) – dfnu Mar 18 '19 at 11:47
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1@Matteo Using the process of those is fine. I only meant exclusion of direct application of them. – Riddle Aaron Mar 18 '19 at 11:51
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Ok, how about you try to use the Generalized MVT (or Cauchy Theorem) to the functions $f(x)$ and $g(x) = x^2$? – dfnu Mar 18 '19 at 11:53
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1@Matteo Thanks!! It works with Cauchy Theorem – Riddle Aaron Mar 18 '19 at 12:02
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1@RiddleAaron Great! Answer the question yourself, and you can close it. – Sarvesh Ravichandran Iyer Mar 18 '19 at 12:02
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post the answer, then: it will be probably useful to others, too – dfnu Mar 18 '19 at 12:03
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@Matteo Wait... I'm not certain that whether I misused squeeze theorem. Please check it after I post an answer. – Riddle Aaron Mar 18 '19 at 12:13
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@Matteo Never mind! Someone posted it already – Riddle Aaron Mar 18 '19 at 12:14
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Ok hope I can be of help... – dfnu Mar 18 '19 at 12:14
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Proof using MVT: let $g(x)=f(x)-\frac 1 2 x^{2}f''(0)$. Then $g''(0)=0$. If we prove the result for $g$ then result for $f$ follows immediately. Now $\frac {g(x)-g(0)} {x^{2}}=\frac {g'(\xi_x)} {x} $for some $\xi_x$ between $0$ and $x$. But $\frac {g'(\xi_x)} {x} =\frac {g'(\xi_x)} {\xi_x} \frac {\xi_x} x \to 0$ because $\frac {\xi_x} x$ is bounded.
Kavi Rama Murthy
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1@RiddleAaron recall that all boils down to Rolle's theorem. Both MVT and GMVT use it with different auxiliary functions. It's just a "game" of going back where you want to go. In the end up to the completeness principle and the definition of limits... – dfnu Mar 18 '19 at 12:17
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