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?

There are several steps that are not clearly allowed.
The first step is okay:
$$
\frac{ \arctan \sqrt{f(x)} - \arctan f(x) }{ \left( \arcsin \sqrt{f(x)} - \arcsin f(x) \right)^2 }
= \frac{ (\arctan \sqrt{f(x)} - \arctan 1) - (\arctan f(x) - \arctan 1) }{ \left( (\arcsin \sqrt{f(x)} - \arcsin 1) - (\arcsin f(x) - \arcsin 1) \right)^2 } \\
= \frac{1}{x-1} \frac{ \frac{\arctan \sqrt{f(x)} - \arctan 1}{x-1} - \frac{\arctan f(x) - \arctan 1}{x-1} }{ \left( \frac{\arcsin \sqrt{f(x)} - \arcsin 1}{x-1} - \frac{\arcsin f(x) - \arcsin 1}{x-1} \right)^2 } \\
$$
Then you assume not only that the limit can be distributed over the internal terms, which would give derivatives at $x=1$, e.g. $\left[ \arctan\sqrt{f(x)} \right]'_{x=1},$ but even that you can take the limits of the derivatives, $\lim_{a \to 0} \left[ \arctan\sqrt{f(x)} \right]'_{x=a}.$
Also, as gimusi has already pointed out, $\arcsin \sqrt{f(x)}$ and $\arcsin f(x)$ are not differentiable when $f(x)=1.$
