Definitionally prove that $\lim_{x \to 0}\frac{f(x)-f(0)}{x^2} = \frac{f''(0)}{2}$ $$\lim_{x \to 0}\frac{f(x)-f(0)}{x^2} = \frac{f''(0)}{2}\quad (f'(0) = 0)$$ 
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?

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.
