Question on l'Hospital's rule Let $f:\textbf{R} \rightarrow \textbf{R}$ be twice differentiable. Suppose that $f''(0) \not= 0$. By MVT it follows that:
$\forall x >0 $ there exists some $  \lambda(x) \in (0,1)$ such that $f(x)-f(0)=xf^{\prime}(x\lambda(x))$
Show that 
$$\displaystyle\lim_{x\rightarrow0}\frac{f(x)-f(0)-xf^{\prime}(0)}{x^2} = \frac{f^{\prime\prime}(0)}{2}$$
and hence that:
$$\displaystyle\lim_{x\rightarrow0}\lambda(x) = \frac{1}{2}$$
I have tried substituting the given equation and using l'Hospital's rule, however all attempts have lead to failures.
 A: With the due correction as pointed out by Mathcounterexamples in the first comment, we get
using the Maclaurin series of order $1$ for $\;f\;$ (with Lagrange remainder) :
$$f(x)=f(0)+f'(0)x+\frac{f''(x\lambda(x))x^2}{2}\implies f(x)-f(0)=f'(0)x+\frac{f''(x\lambda(x))x^2}{2}\implies$$
$$\frac{f(x)-f(0)-xf'(0)}{x^2}=\frac{f''(x\lambda(x))}2\xrightarrow[x\to0]{}\frac{f''(0)}2 \tag{1}$$
and on the other hand
$$\frac{f(x)-f(0)-xf'(0)}{x^2}=\frac{xf'(x\lambda(x))}{x^2}-\frac{f'(0)}x=\frac{f'(x\lambda(x))-f'(0)}x=$$
$$=\frac{f'(x\lambda(x))-f'
(0)}{x\lambda(x)}\cdot\lambda(x)\xrightarrow[x\to0]{}f''(0)\lambda(0) \tag{2}$$
The above is assuming the limit of $\;\lambda(x)\;$ when $\;x\to0\;$ exists.
A: The key fact is that $f''(0)\ne0$. The Inverse Function Theorem then gives you that $f'$ is invertible on some neighbourhood of $0$, so there exists $g$ with $g(f'(x))=x$ and
$$
g'(f'(0))=\frac1{f''(g(f'(0)))}=\frac1{f''(0)}.
$$ So there exists $g$, differentiable on some interval $I\ni0$, with $g(f'(x))=x$. From $f(x)-f(0)=xf'(x\lambda(x))$ we get, for $x\in I$, 
$$
\lambda(x)=\frac{g\left(\frac{f(x)-f(0)}x\right)}x=\frac{g\left(\frac{f(x)-f(0)}x\right)-g(f'(0))}x=\frac{g\left(\frac{f(x)-f(0)}x\right)-g(f'(0))}{\frac{f(x)-f(0)}x-f'(0)}\,\frac{ {f(x)-f(0)}-xf'(0)}{x^2}.
$$
Using the Mean Value Theorem, 
$$
\lambda(x)=g'(c(x))\,\frac{ {f(x)-f(0)}-xf'(0)}{x^2}
$$
with $c(x)$ between $f'(0)$ and $\frac{f(x)-f(0)}x$. Then 
$$
\lim_{x\to0}\lambda(x)=g'(f'(0))\,\frac{f''(0)}2=\frac1{f''(0)}\,\frac{f''(0)}2=\frac12.
$$
A: Let us compute the limit
$$
\lim_{x\to 0} \frac{f(x) - f(0) - x f'(0)}{x^2}\,.
$$
By de l'Hopital's rule, this limit is equal to
$$
\lim_{x\to 0} \frac{f'(x) - f'(0)}{2x}\,,
$$ 
provided that this second limit exists. On the other hand, by the definition of derivative, this second limit equals $f''(0)/2$.
Now, let $\lambda(x)\in (0,1)$ be such that $f(x) - f(0) = x f'(x \lambda(x))$.
We have that
$$
g(x) := \frac{f(x) - f(0) - x f'(0)}{x^2}=
\frac{f'(x\lambda(x)) - f'(0)}{x}=
\frac{f'(x\lambda(x)) - f'(0)}{x\lambda(x)}\, \lambda(x)=: \varphi(x)\, \lambda(x).
$$
Clearly, by definition of derivative,
$$
\lim_{x\to 0}\varphi(x) = 
\lim_{x\to 0} \frac{f'(x\lambda(x)) - f'(0)}{x\lambda(x)} = f''(0) \neq 0,
$$
whereas, by the previous step, $\lim_{x\to 0} g(x) = f''(0)/2$.
Finally,
$$
\lim_{x\to 0}\lambda (x) = \lim_{x\to 0} \frac{g(x)}{\varphi(x)} = \frac{1}{2}.
$$
