Complex limit involving roots of unity Suppose $\eta$ is a complex number satisfying $\eta^r = 1$. Let $g(t), k(t)$ be two functions of complex variable $t$ such that $g \neq k$ but $g(\eta) = k(\eta)$. We want to evaluate the limit
$$
\lim_{t \to \eta} \frac{t^rg(t) - k(t)}{(t^r + 1)g(t) - 2k(t)}.
$$
Intuitively, I believe this limit equals $\frac{1}{2}$. This is immediately seen, for example, when $g(\eta) \neq k(\eta)$. However, I have not been able to show this in the present case (i.e. when $g(\eta) = k(\eta)$).
I have tried rearranging, L'Hopital's rule, subtracting $1/2$ and sandwiching the difference, to no avail. I feel like I'm missing something elementary?
EDIT: As per the comments, this seems to be false in general. However, I am also interested in the special case when $g$ and $k$ are monic polynomials of the same degree. Perhaps this is enough to find the limit?
Thanks!
 A: If you assume that $g$ and $k$ are analytic at and around $\eta$ then both have a power series expansion valid in a neighbourhood of $\eta$.  Because $g(\eta)=k(\eta)$ but $g$ and $k$ are not equal, we know the power series are identical up to the $n$th term and differ in the $n+1$th term for some $n \geqslant 0$.  Then we have,
\begin{align}g(t) &= a_0+a_1(t-\eta)+a_2(t-\eta)^2+\cdots \\
k(t) &= a_0+a_1(t-\eta)+\cdots+a_n(t-\eta)^n+b_{n+1}(t-\eta)^{n+1}+\cdots 
\end{align}
and as $t \to\eta$,
\begin{align}
\frac{t^r g(t)-k(t)}{(t^r+1)g(t)-2k(t)} &=
\frac{\frac{t^r-1}{t-\eta}a_0 + (t-\eta)^n (t^r a_{n+1}-b_{n+1}) + o((t-\eta)^n)}
{\frac{t^r-1}{t-\eta}a_0+(t-\eta)^n ((t^r+1) a_{n+1}-2b_{n+1}) + o((t-\eta)^n)}.
\end{align}
We can now distinguish three cases:  (i) if $a_0=0$ the limit is $1/2$ for all $n \geqslant 0$; (ii) if $a_0 \neq 0$ and $n > 0$ the limit is $1$; and (iii), if $a_0 \neq 0$ and $n=0$ the limit is,
\begin{align}
\frac{r\eta^{r-1}a_0 + a_1-b_1}{r\eta^{r-1}a_0+2(a_1-b_1)}.
\end{align}
