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?

Probably the derivation is wrong because arcsin x is not differentiable at $x=1$, therefore your first step is not allowed.
As an alternative, since $f(x)$ is continuos at $x=1$ we have that
$$L = \lim_{y \to 1^-} {\frac{\arctan{\sqrt{y}-\arctan{y}}}{ \left (\arcsin{\sqrt{y}}-\arcsin{y}\right)^2}}$$
then we can use that


*

*$\arcsin(x) = \arctan\left(\frac{x}{\sqrt{1 - x^2}}\right)$

*$\arctan(u) - \arctan(v) = \arctan\left(\frac{u - v}{1 +
   uv}\right) $


to obtain
$$L = \lim_{y \to 1^-} 
\frac{ \arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)}
{ \left[\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} { \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\right]^2 }=\frac{3+ 2\sqrt 2}{4}=\left( \frac{\sqrt{2}+1}{2}\right)^2$$
indeed by $y=1-u$ with $u\to 0^+$ by binomial expansion $\sqrt y = 1-\frac12 u +o(u)$ it easy to show that


*

*$\arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)\sim \frac 14 u$

*$\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} {
   \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\sim (\sqrt 2 -1) \sqrt u$


and therefore
$$\frac{ \arctan\left( \frac{\sqrt y -y}{1+y\sqrt y}\right)}
{ \left[\arctan \left(\frac{\sqrt y \sqrt{1-y^2}-y\sqrt{1-y}} { \sqrt{(1-y)(1-y^2)}+y\sqrt y} \right)\right]^2 }\sim \frac{\frac 14 u}{(\sqrt 2 -1)^2  u}=\frac{3+ 2\sqrt 2}{4}$$
