prove: $\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} = 1$ $f(x)$ and $g(x)$ both differentiable twice at $x = a$ and we know that $f''(a) =g''(a)+f(a)$, $f(a) = g(a) = f'(a) = g'(a) \not = 0$
(we don't know if $f(x)$ and $g(x)$ are differentiable anywhere else except $x=a$)
prove: $\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} = 1$
I tried to substitute $f^2(x)$ and $g^2(x)$ with a square of taylor expansion of degree 1 (with remainder of degree 2) for example: 
$$f^2(x)= (f(a) + f'(a)x + \frac{f''(\xi)}{2}x^2)^2$$
same for $g^2(x)$.
Now because $f''(x)$ is continuous and since $x \rightarrow a$ we can conclude that $f(\xi) = f(a)$.
Now after simplifying the original expression i get: 
$$\lim\limits_{x \rightarrow a} \frac{f^2(a)(3x^2+2x) + f(a)g''(a) \frac{x^4}{2}}{f^2(a) \cdot (x^2 + \frac{x^2}{2}) + f(a)(x+\frac{x^2}{2})g''(a)x^2 + g''}$$
 A: Let's let $L\not=0$ denote the common value of $f(a)=g(a)=f'(a)=g'(a)$. Note that $f^2(x)-g^2(x)=(f(x)+g(x))(f(x)-g(x))$, so it suffices to show that
$$\lim_{x\to a}{f(x)-g(x)\over(f(x)-f(a))^2}={1\over2L}$$
since $\lim_{x\to a}(f(x)+g(x))=2L\not=0$.  Invoking L'Hopital, it suffices to show
$$\lim_{x\to a}{f'(x)-g'(x)\over2f'(x)(f(x)-f(a))}={1\over2L}$$
which, since $2f'(x)\to2f'(a)=2L\not=0$, reduces to showing
$$\lim_{x\to a}{f'(x)-g'(x)\over f(x)-f(a)}=1$$
If we knew that $f$ and $g$ are both twice differentiable in a neighborhood of $x=a$ (and if the second derivatives are continuous at $a$), then we could invoke L'Hopital again.  But it turns out we don't need to invoke L'Hopital at all; we can instead use the definition of derivative:  Using the fact that $f'(a)=g'(a)$, we have
$${f'(x)-g'(x)\over f(x)-f(a)}={(f'(x)-f'(a))-(g'(x)-g'(a))\over f(x)-f(a)}={{f'(x)-f'(a)\over x-a}-{g'(x)-g'(a)\over x-a}\over {f(x)-f(a)\over x-a}}$$
hence
$$\lim_{x\to a}{f'(x)-g'(x)\over f(x)-f(a)}={\lim_{x\to a}{f'(x)-f'(a)\over x-a}-\lim_{x\to a}{g'(x)-g'(a)\over x-a}\over\lim_{x\to a} {f(x)-f(a)\over x-a}}={f''(a)-g''(a)\over f'(a)}={(g''(a)+L)-g''(a)\over L}=1$$
as desired.
A: Since $f(a)=g(a)$, numerator en denominator tend to $0$ for $x \to a$. Apply l'Hôpital's rule:
$$\lim_{x \to a} \frac{f^2(x)-g^2(x)}{(f(x)-f(a))^2} =
\lim_{x \to a} \frac{2f(x)f'(x)-2g(x)g'(x)}{2(f(x)-f(a))f'(x)} = 
\lim_{x \to a} \frac{f(x)f'(x)-g(x)g'(x)}{f(x)f'(x)-f(a)f'(x)}$$
Since also $f(a)=g(a)=f'(a)=g'(a)$, you again have "$0/0$" so apply l'Hôpital's rule again:
$$\begin{array}{rl}
= & \displaystyle \lim_{x \to a} \frac{f(x)f''(x)+f'(x)^2-g(x)g''(x)-g'(x)^2}{f(x)f''(x)+f'(x)^2-f(a)f''(x)} \\[7pt]
= & \frac{\displaystyle \lim_{x \to a} \left( f(x)f''(x)+f'(x)^2-g(x)g''(x)-g'(x)^2 \right) }{\displaystyle \lim_{x \to a} \left( f(x)f''(x)+f'(x)^2-f(a)f''(x)\right)} \quad \color{green}{(*)}
\end{array}
$$
Now with $\color{blue}{f''(a) = g''(a)+f(a)}$ and $\color{red}{f(a)=g(a)=g'(a)}$, the numerator becomes:
$$f(a)\left( \color{blue}{g''(a)+f(a)} \right)+f'(a)^2-g(a)g''(a)-g'(a)^2 \\
= \color{red}{g(a)}\left( \cancel{\color{blue}{g''(a)}}+\bcancel{\color{red}{g(a)}} \right)+f'(a)^2\require{cancel}\cancel{-g(a)g''(a)}\bcancel{-\color{red}{g(a)^2}} \\
= f'(a)^2$$
and the denominator becomes:
$$\require{cancel}
\cancel{f(a)f''(a)}+f'(a)^2\cancel{-f(a)f''(a)} = f'(a)^2$$
So:
$$\color{green}{(*)} = \frac{f'(a)^2}{f'(a)^2} = 1$$
Caveat: this method requires continuity of $f''$ at $a$, so that $\lim_{x \to a}f''(x) = f''(a)$.
A: If you use the Taylor polynomial, the answer is easy and short. Note
$$ f(x)=f(a)+f'(a)(x-a)+\frac{1}{2}f''(\xi)(x-a)^2, g(x)=g(a)+g'(a)(x-a)+\frac{1}{2}g''(\eta)(x-a)^2$$
where $\xi$ and $\eta$ are between $a$ and $x$.
So
\begin{eqnarray}
\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} &=&\lim\limits_{x \rightarrow a} (f(x)+g(x))\frac{f(x)- g(x)}{(f(x) -f(a))^2}\\
&=&\lim\limits_{x \rightarrow a} 2f(a)\frac{\frac{1}{2}(f''(\xi)- g''(\eta))(x-a)^2}{[f'(a)(x-a)+O((x-a)^2)]^2}\\
&=&\lim\limits_{x \rightarrow a} f(a)\frac{f''(\xi)- g''(\eta)}{[f'(a)+O((x-a))]^2}\\
&=&f(a)\frac{f(a)}{f^2(a)}\\
&=&1.
\end{eqnarray}
A: We can use Taylor series also (instead of L'Hospital's Rule). We have
\begin{align}
f(a + h) &= f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) + o(h^{2})\notag\\
g(a + h) &= g(a) + hg'(a) + \frac{h^{2}}{2}g''(a) + o(h^{2})\notag
\end{align}
We can proceed as follows
\begin{align}
L &= \lim_{x \to a}\frac{f^{2}(x) - g^{2}(x)}{(f(x) - f(a))^{2}}\notag\\
&= \lim_{x \to a}\{f(x) + g(x)\}\cdot\frac{f(x) - g(x)}{(f(x) - f(a))^{2}}\notag\\
&= \{f(a) + g(a)\}\lim_{x \to a}\dfrac{f(x) - g(x)}{\left(\dfrac{f(x) - f(a)}{x - a}\right)^{2}(x - a)^{2}}\notag\\
&= \frac{2f(a)}{\{f'(a)\}^{2}}\lim_{x \to a}\frac{f(x) - g(x)}{(x - a)^{2}}\notag\\
&= \frac{2}{f(a)}\lim_{h \to 0}\frac{f(a + h) - g(a + h)}{h^{2}}\text{ (putting }x = a + h)\notag\\
&= \frac{2}{f(a)}\lim_{h \to 0}\frac{1}{h^{2}}\cdot\{f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) - g(a) - hg'(a) - \frac{h^{2}}{2}g''(a) + o(h^{2})\}\notag\\
&= \frac{2}{f(a)}\cdot\frac{f''(a) - g''(a)}{2}\notag\\
&= \frac{2}{f(a)}\cdot\frac{f(a)}{2}\notag\\
&= 1\notag
\end{align}
