How would you calculate a derivative of $ f(x)= \frac{\sqrt{x+1}}{2-x}$ by the limit definition? I have a function defined as follows:
$$
f(x)= \frac{\sqrt{x+1}}{2-x}
$$
I tried to calculate the derivative using the limit definition using four methods, but I was unsuccessful in any. Could someone help me calculate it and explain the method?
$$
1) \lim_{h\to 0} =\frac{\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}}h
$$
$$
2)\lim_{z\to x} =\frac{\frac{\sqrt{z+1}}{2-z}-\frac{\sqrt{x+1}}{2-x}}{z-x}
$$
$$
3)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u=\sqrt{x+1}
$$
$$
\lim_{h\to 0} =\frac{\frac{u+h}{3-(u+h)^2}-\frac{u}{3-u^2}}h
$$
$$
4)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u={x+1};
$$
$$
\lim_{h\to 0} =\frac{\frac{\sqrt{u+h}}{3-(u+h)}-\frac {\sqrt{u}}{3-u}}h
$$

Note that we have
$$\begin{align}
\frac{\sqrt{x+1+h}}{2-x-h}-\frac{\sqrt{x+1}}{2-x}&=\frac{\sqrt{x+1+h}-\sqrt{x+1}}{2-x-h}+\frac{h\sqrt{x+1}}{(2-x-h)(2-x)}\\\\
&=\frac{h}{(2-x-h)(\sqrt{x+1+h}+\sqrt{x+1})}+\frac{h\sqrt{x+1}}{(2-x-h)(2-x)}\\\\
\end{align}$$
Now divide by $h$ and let $h\to 0$ to find
$$\frac{d}{dx}\left(\frac{\sqrt{x+1}}{2-x}\right)=\frac{1}{2(2-x)\sqrt{x+1}}+\frac{\sqrt{x+1}}{(2-x)^2}$$
