Finding $ \lim\limits_{x \to 0} \frac{f(x) - f(0) - x f^\prime(x)}{x^2} $ I'm trying to find $ \lim\limits_{x \to 0} \dfrac{f(x) - f(0) - x f^\prime(x)}{x^2} $ with $ f: \mathbb{R} \rightarrow \mathbb{R} $ a twice differentiable function.
I noticed that $ \dfrac{f(x) - f(0) - x f^\prime(x)}{x^2} = \dfrac{f(x) - (f(0) + x f^\prime(x)}{x^2} $ as $ f(0) + x f^\prime(0) $ being the equation of the the tangent line at $0$. But I have no idea on how to take advantage to this information.
Any input is appreciated.
 A: For $x$ near $0$, we have $$\frac{f(x) -f(0) -x f^{\prime}(x)}{x^{2}} = \frac{f(x) -f(0) -x f^{\prime}(0)}{x^{2}} -\frac{f^{\prime}(x) -f^{\prime}(0)}{x} \, \text{.}$$ As $f$ is twice differentiable at $0$, we have $$f(x) = f(0) +x f^{\prime}(0) +\frac{f^{\prime \prime}(0)}{2} x^{2} +o_{0}\left (x^{2} \right)$$ and $$f^{\prime}(x) = f^{\prime}(0) +x f^{\prime \prime}(0) +o_{0}(x) \, \text{.}$$ Therefore, we have $$\frac{f(x) -f(0) -x f^{\prime}(x)}{x^{2}} \underset{x \rightarrow 0}{\longrightarrow} \frac{f^{\prime \prime}(0)}{2} -f^{\prime \prime}(0) = -\frac{f^{\prime \prime}(0)}{2} \, \text{.}$$
A: If $f$ is twice differentiable, say at $0$, then there exists a function $h_2: \mathbb{R} \to \mathbb{R}$ such that $\lim\limits_{x \to 0} h_{2}(x)=0$, and 
$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+h_{2}(x)x^2$$. You can finish it now!
So $\lim\limits_{x \to 0 } \frac{f(x)-f(0)-xf'(0)}{x^2}=\frac{f''(0)}{2}$.
Now write:
$$\frac{f(x)-f(0)-xf'(x)}{x^2}=\frac{f(x)-f(0)-xf'(0)-x(f'(x)-f'(0))}{x^2}=\frac{f(x)-f(0)-xf'(x)}{x^2}-\frac{f'(x)-f'(0)}{x}$$
It should not be hard to see that the limit is $-\frac{1}{2}f''(0)$, by just splitting the two terms.
A: An intuitive answer
As with your idea of a tangent line, we know that, for $x$ small, $$f'(x)\approx f'(0)+xf''(0).$$
Integrating, we get
$$f(x)\approx f(0)+xf'(0)+\frac{x^2}{2}f''(0).$$
Substitute into the formula and you obtain 
$$-\frac{1}{2}f''(0).$$
A: Rewrite the function to
$\frac{\frac{f(x)-f(0)}{x} - f^\prime(x)}{x}$
Notice that the nominator and denominator both tend to 0 as x tends to 0 (since $f^\prime(x)$ is continuous), so we can use L'Hopitals rule to find the limit as the limit of
$\frac{f^\prime(x)x-f(x) + f(0)}{x^2}-f^{''}(x)$ 
Notice that the first term is the original function time -1, so we get that the limit is $-\frac{1}{2}f^{''}(x)$
