<s>[INST] If $y=\sin{(m\cos^{-1}\sqrt{x})}$ then prove that $\lim\limits_{x\to 0}\frac{y_{n+1}}{y_n}=\frac{4n^2-m^2}{4n+2}$ If $y=\sin{(m\cos^{-1}\sqrt{x})}$ then prove that $$\lim\limits_{x\to 0}\frac{y_{n+1}}{y_n}=\frac{4n^2-m^2}{4n+2}$$
Note: $y_n = D^ny(x)$. To use: Libnitz rule of successive Differentiation. 
Attempt:
$y_1=-1/2\,{\dfrac {m\cos \left( m\arccos \left( \sqrt {x} \right)  \right) 
}{\sqrt {x}\sqrt {1-x}}}$ then 
$4y_1^2x(1-x)=m^2(1-y^2)$
differentiating n times
$$8y_1y_{n+1}x(1-x)+n4(1-2x)2y_1y_n+\frac{n(n-1)}{2}(-8)2y_1y_{n-1}=-2m^22yy_n$$
Problem is here $y_{n+1}, y_n, y_{n-1}$ are all present. What to do to prove the result.
 [/INST] We have $y= \sin(m \cos ^{-1} \sqrt{x})$ & using the notation $y_n=\frac{d^n y}{dx^n}$. You start fine ... differentiating (multiple use of the chain rule) squaring & using Pythagorus to get to
\begin{eqnarray*}
4x(1-x)y_1^{2}=m^2(1-y^2)
\end{eqnarray*}
Now differentiate this
\begin{eqnarray*}
8x(1-x)y_1 y_2 +(4-8x)y_1^{2}+2 m^2y_1 y=0
\end{eqnarray*}
Now cancel $2y_1$ and keep differentiating
\begin{eqnarray*}
4x(1-x) y_2 +2(1-2x)y_1+ m^2 y=0 \\
4x(1-x) y_3 +6(1-2x)y_2+ (m^2-4) y_1 =0 \\
4x(1-x) y_4 +10(1-2x)y_3+ (m^2-16) y_2=0 \\
\end{eqnarray*}
It is easy to show by induction that
\begin{eqnarray*}
4x(1-x) y_{n+2} +2(2n+1)(1-2x)y_{n+1}+ (m^2-4n^2) y_n=0. \\
\end{eqnarray*}
Now substiute $x=0$ and the required result follows.
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