When does this limit coincide with the derivative of the function? I am working with the following limit
$$\lim_{h\to 0}\frac{f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)}{h}$$
where $c$ is a constant with $0 \leq c \leq 1$. Note that $$\lim_{h \to 0}\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}=\frac{1}{c+1}$$
Under what conditions on the function f will this limit coincide with the derivative of $f$ at $\frac{1}{1+c}$ (assuming this exists):
$$f'\left(\frac{1}{1+c}\right)=\lim_{h \to 0}\frac{f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{c+1}\right)}{h}$$
 A: Assume that $f \in C^2$ so that we can use taylor's formula without concerns. We want to establish conditions on $\varepsilon(h)$ so that
$$
f'(a)=\lim_{h \to 0}\dfrac{f(a+h) - f(a+\varepsilon(h))}{h}
$$
Using Taylor's formula,
\begin{align*}
\dfrac{f(a+h) - f(a+\varepsilon(h))}{h}=&\frac{1}{h}\left(f(a)+f'(a) h +\frac{f''(\xi_1)}{2} h^2 -f(a)-f'(a) \varepsilon(h) - \frac{f''(\xi_2)}{2} \varepsilon(h)^2 \right)\\
= & f'(a)-f'(a) \frac{\varepsilon(h)}{h}+\frac{f''(\xi_1)}{2} h -\frac{f''(\xi_2)}{2} \frac{\varepsilon(h)^2}{h}
\end{align*}
Making $h \to 0$,
$$
\lim_{h \to 0} \dfrac{f(a+h) - f(a+\varepsilon(h))}{h} = f'(a)-f'(a)\cdot \lim_{h\to 0}\frac{\varepsilon(h)}{h} - \lim_{h\to 0} \frac{f''(\xi_2)}{2} \frac{\varepsilon(h)^2}{h}
$$
So you see that if $\dfrac{\varepsilon(h)}{h} \to 0$, the proposed limit coincides with $f'(a)$.
A: By the MVT, there is $\xi$ between $\frac{1}{c+1}+h$ and $\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}$ such that
\begin{eqnarray}
&&f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)\\
&=&f'(\xi)\left[\left(\frac{1}{c+1}+h\right)-\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)\right].
\end{eqnarray}
Let
$$ g(h)=\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1)}}=\frac{1}{2+\sqrt{c\left(\frac{1}{\frac{1}{c+1}+h}-1\right)}(1-c^{-1)}} $$
and then $g(0)=\frac{1}{c+1}$ and
$$ g'(0)=-\frac{1}{\left[2+\sqrt{c\left(\frac{1}{\frac{1}{c+1}+h}-1\right)}(1-c^{-1)}\right]^2}\cdot\frac12 \frac{1-c^{-1}}{\sqrt{c\left(\frac{1}{\frac{1}{c+1}+h}-1\right)}}(-1)\frac{1}{(\frac{1}{c+1}+h)^2}\bigg|_{h=0}=\cdots$$
So
$$ g(h)=g(0)+g'(0)h+\cdots $$
and hence
$$\lim_{h->0}\frac{f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)}{h}=-f'(\frac1{c+1})\dots.$$
