Convergence of fixed points as a consequence of pointwise convergence? Setup: Let $p := p(n)$ be such that $\lim_{n \to \infty} pn = \lambda > 0$, and suppose $f_n(t) := (1 + (t-1)p)^n$, with domain $[0, 1]$.
Goal: I am trying to show that if $\theta_n$ is a fixed point of $f_n$:
$$
\theta_n = f_n(\theta_n), \quad \text{for all}~n,
$$
then  $\theta_n$ converges  and its limit $\theta$ solves $e^{\lambda(\theta - 1)} = \theta$.
What I tried: I can see that $\theta \mapsto e^{\lambda(\theta - 1)}$ is an increasing function of $\theta$ and so the fixed point is unique. Moreover, for $\epsilon > 0$ and $n$ large enough, we can provide the estimate
$$
(1 - \epsilon)\lambda \leq pn \leq (1 + \epsilon)\lambda. 
$$
Therefore, for $n$ sufficiently large: we have the sandwich relation
$$
\left(1 + \lambda\frac{(t-1) - \epsilon}{n} \right)^n \leq f_n(t) \leq 
\left(1 + \lambda\frac{(t-1) + \epsilon}{n} \right)^n, \quad \mbox{for}~t \in [0, 1].
$$
And so taking limits, $e^{\lambda((t-1) - \epsilon)} \leq \lim_n f_n(t) \leq e^{\lambda((t - 1) + \epsilon)}$, and by taking $\epsilon\downarrow 0$, we see that $f_n \to f$ pointwise on $[0, 1]$, where $f(t) := e^{\lambda (t- 1)}$.
Intuitively, it seems that since $\lambda > 0$, eventually $f_n$ have unique fixed points and so since $f_n \to f$, it seems that $\theta_n \to \theta$ should follow, but I do not know how to show this.
Comment regarding uniform convergence. Suppose that $f_n \to f$ uniformly on $[0, 1]$. Then let $\epsilon > 0$ and define
$$
\delta(\epsilon) := \inf_{t : |\theta - t| \geq \epsilon} |t - f(t)|.
$$
Since $f$ is continuous and $\theta$ is the unique fixed point of $f$, it follows that $\delta(\epsilon) > 0$. Now let $n$ be large enough such that
$\|f_n - f\|_{\infty} < \delta(\epsilon)$. Note that $|\theta_n - \theta| <\epsilon$. (If not, then:
$$
\delta(\epsilon) \leq 
|f(\theta_n) - \theta_n| \leq |f_n(\theta_n) - f(\theta_n)| + |f_n(\theta_n) - \theta_n| < 
\delta(\epsilon) + |f_n(\theta_n) - \theta_n|.
$$
So cancelling terms, we get a contradiction to the fact $\theta_n$ is a fixed point of $f_n$.) Putting the pieces together, we see that for each $\epsilon > 0$, we have $|\theta_n - \theta| <\epsilon$, for sufficiently large $n$, whence $\theta_n \to \theta$.
Consequence: it is sufficient to show that $f_n \to f$ uniformly.
 A: Your sandwich argument estimate is already essentially uniform. Namely,
$$e^{\lambda((t-1)+\epsilon)} - e^{\lambda((t-1)-\epsilon)} = e^{\lambda(t-1)} (e^{\lambda \epsilon} - e^{-\lambda \epsilon}).$$
As $\epsilon \to 0$, the right-hand-side becomes $1-1=0$.
Edit: Some more details. There's probably a cleaner way to arrange it.
Let
\begin{align*}
  L_{\epsilon, n}(t) &:= (1+\lambda \frac{(t-1)-\epsilon}{n})^n, \\ 
  U_{\epsilon, n}(t) &:= (1+\lambda \frac{(t-1)+\epsilon}{n})^n, \\ 
  L_\epsilon(t) &:= \exp(\lambda((t-1)-\epsilon), \\
  U_\epsilon(t) &:= \exp(\lambda((t-1)+\epsilon).
\end{align*}
For each fixed $\epsilon>0$, we have $L_{\epsilon, n}(t) \to L_\epsilon(t)$ uniformly as $n \to \infty$, since
\begin{align*}
  \log\left(1+\lambda\frac{(t-1)-\epsilon}{n}\right)^n
   &= n\log\left(1+\lambda\frac{(t-1)-\epsilon}{n}\right) \\
   &= n\left(\lambda\frac{(t-1)-\epsilon}{n} + O\left(\left(\lambda\frac{(t-1)-\epsilon}{n}\right)^2\right)\right) \\
   &= \lambda(t-1)-\epsilon + O(1/n).
\end{align*}
Likewise $U_{\epsilon, n}(t) \to U_\epsilon(t)$ uniformly as $n \to \infty$. As noted, $U_\epsilon(t) - L_\epsilon(t) \to 0$ uniformly as $\epsilon \to 0$.
We have
\begin{align*}
  |f_n - f|
  &\leq |f_n - L_{\epsilon, n}| + |L_{\epsilon, n} - L_\epsilon| + |L_\epsilon - f| \\
  &\leq |U_{\epsilon, n} - L_{\epsilon, n}| + |L_{\epsilon, n} - L_\epsilon| + |L_\epsilon - U_\epsilon| \\
  &\leq |U_{\epsilon, n} - U_\epsilon| + |U_\epsilon - L_\epsilon| + |L_\epsilon - L_{\epsilon, n}| + |L_{\epsilon, n} - L_\epsilon| + |L_\epsilon - U_\epsilon|.
\end{align*}
For each $\delta>0$, there is some $\epsilon>0$ such that $|U_\epsilon - L_\epsilon| < \delta$, and for that $\epsilon$, for large enough $n$ we have $|L_{\epsilon, n} - L_\epsilon|, |U_{\epsilon, n} - U_\epsilon| < \delta$. Hence $f_n \to f$ uniformly on $[0, 1]$.
This last sequence of inequalities feels redundant, but I couldn't quickly find a cleaner way through, and this works.
