Finding the limit by using the definition of derivative. Here is the problem.

Let $f$ be the function that has the value of $f(1)=1$ and $f'(1)=2$. Find the value of
  $$ L = \lim_{x \to 1} {\frac{\arctan{\sqrt{f(x)}-\arctan{f(x)}}}{ \left (\arcsin{\sqrt{f(x)}}-\arcsin{f(x)}\right)^2}} $$

I have tried using
$$
L=\lim_{x\to 1}
\frac{1}{x-1}\frac{\frac{\arctan{\sqrt{f(x)}}-\arctan{\sqrt{f(1)}}}{x-1}-\frac{\arctan{{f(x)}}-\arctan{{f(1)}}}{x-1}}
{\left [\frac{\arcsin{\sqrt{f(x)}}-\arcsin{\sqrt{f(1)}}}{x-1}-\frac{\arcsin{{f(x)}}-\arcsin{{f(1)}}}{x-1}  \right ]^2}
$$
and reduced that big chunks by using $f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$ which I got
$$
\begin{split}
L&=\lim_{a\to 1}
\frac{1}{a-1}\frac{\Big [ \arctan\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arctan{f(x)} \Big ]'_{x=a}}
{\Big [ \arcsin\sqrt{f(x)} \Big ]'_{x=a} - \Big [ \arcsin{f(x)} \Big ]'_{x=a}}\\[2em]
&=\lim_{a\to 1}
\frac{1}{a-1} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{1+(f(a))^2}f'(a)}
{\left [  \frac{1}{\sqrt{1-f(a)}}\frac{1}{2\sqrt{f(a)}}f'(a)-\frac{1}{\sqrt{1-(f(a))^2}}f'(a) \right ]^2}\\[2em]
&=\lim_{a\to 1}
\frac{1}{f'(a)\frac{a-1}{1-f(a)}} \frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [  \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}\\[2em]
&=\lim_{a\to 1}{-\frac{\frac{1}{1+f(a)}\frac{1}{2\sqrt{f(a)}}-\frac{1}{1+(f(a))^2}}
{\left [  \frac{1}{2\sqrt{f(a)}}-\frac{1}{\sqrt{1+f(a)}} \right ]^2}}\\[2em]
&=\boxed{(\sqrt{2}+1)^2}
\end{split}
$$
but the answer keys tell me that the answer of this problem is $L=\left( \frac{\sqrt{2}+1}{2}\right)^2$.
So, Can someone please explain to me what did I do wrong?

We shall need several times the following principle: If $f(0)=0$, $\>f'(0)=1$, and $\lim_{x\to0} g(x)=0$, then we may write 
$$f\bigl(g(x)\bigr)=g(x)\>h(x)\ ,\tag{1}$$ whereby the function $x\mapsto h(x)$ defined by $(1)$ satisfies $\lim_{x\to0}h(x)=1$.
We only need that $f(x)\nearrow1$ when $x\nearrow 1$ (note that your expression is undefined when $x>1$). There are functions $x\mapsto \alpha:=\alpha(x)$ and $x\mapsto\beta:=\beta(x)$ with $$0<\alpha<\beta\>,\qquad  \lim_{x\to1-}\alpha(x)=\lim_{x\to1-}\beta(x)=0\ ,$$ such that $$\sqrt{f(x)}=\cos\alpha\>,\qquad f(x)=\cos\beta\ .$$
This ensures
$$\arcsin\sqrt{f(x)}-\arcsin f(x)=\beta-\alpha\ .$$
On the other hand
$$\arctan\sqrt{f(x)}-\arctan f(x)=\arctan{\cos\alpha-\cos\beta\over 1+\cos\alpha\cos\beta}={1\over2}(\cos\alpha-\cos\beta)\> g(x)\>,$$
whereby $\lim_{x\to1-}g(x)=1$. Now
$${1\over2}(\cos\alpha-\cos\beta)=\sin{\beta+\alpha\over2}\sin{\beta-\alpha\over2}={1\over4}(\beta+\alpha)(\beta-\alpha) h(x)\ ,$$ 
whereby $\lim_{x\to1-} h(x)=1$. It follows that
$$L={1\over4}\lim_{x\to1-}{\beta+\alpha\over\beta-\alpha }={1\over4}\lim_{x\to1-}{1+{\alpha\over\beta}\over1-{\alpha\over\beta}}.$$
As $\cos^2\alpha=\cos\beta$ for all $x<1$ we have
$\>\sin^2\alpha=2\sin^2{\beta\over2}$,
which then leads to $\lim_{x\to1-}{\alpha\over\beta}={1\over\sqrt{2}}$, so that we end up with
$$L={1\over4}\bigl(3+2\sqrt{2}\bigr)\ .$$
