Does this nonlinear equation give eigenvectors for matrix $A$? Let $A$ be a matrix with nonnegative entries and $y$ be a vector satisfying $$\sum_{k = 1}^\infty y^k =  Ay $$ where $[y^k]_i = y_i^k$ is the pointwise exponential and $0 < y_i < 1$. Is $y$ an eigenvector of $A$? Then proof shown in the text I am reading now argues as follows.

Let $y = \epsilon x $ where $\epsilon > 0$ is arbitrarily small. Dividing the above equation by $\epsilon$ gives $$Ax = x + \epsilon x^2 + O(\epsilon^2)$$
For sufficiently small $\epsilon > 0$, it reduces to $Ax = x$.

My question is how can we take $\epsilon \to 0$? The vector $x$ clearly depends on $\epsilon$.
 A: (Not an answer but too long for a comment)
Justin misquote a little the authors; their alledged proof was even worse. What they wrote was (denoting by $y$ their $V_\infty$ and making explicit that it depends on $\varepsilon$ via $\tau=\tau_c+\varepsilon$ where $\tau_c$ is the inverse of the largest eigenvalue of $A$):

For every $\varepsilon>0$, there is a unique non-zero vector $y(\varepsilon)$ such that $$Ay(\varepsilon)=\frac1{\tau_c+\varepsilon}\sum_{k=1}^\infty y^k(\varepsilon).$$
Let $x(\varepsilon):=\frac1\varepsilon y(\varepsilon),$ then (the components of this vector are non-negative and)
$$Ax(\varepsilon)=\frac1{\tau_c+\varepsilon}x(\varepsilon)+\frac\varepsilon{\tau_c+\varepsilon}x(\varepsilon)^2+O(\varepsilon^2).$$
For sufficiently small $\varepsilon>0$, this reduces to
$$Ax(\varepsilon)=\frac1{\tau_c+\varepsilon}x(\varepsilon),$$

which is clearly a bunch of gross nonsense.
A: Assuming that you are not misquoting the authors, the mentioned “proof” is pure nonsense. The vector $y$ is not necessarily an eigenvector of $A$. E.g. we have
$$
\underbrace{\pmatrix{1&3/2\\ 1&0}}_A\ \underbrace{\pmatrix{1/2\\ 1/3}}_y
=\pmatrix{1\\ 1/2}
=\pmatrix{\frac12/(1-\frac12)\\ \frac13/(1-\frac13)}
=\pmatrix{\sum_{k=1}^\infty (\frac12)^k\\ \sum_{k=1}^\infty (\frac13)^k}
$$
but $y$ is not an eigenvector of $A$, because $Ay$ is not a scalar multiple of $y$.
