Assuming that $\lim_{n \rightarrow \infty}f(n)^n$ exists and is finite, is it true that
$$\lim_{n \rightarrow \infty}(f(n) - o(n))^n = \lim_{n \rightarrow \infty}f(n)^n$$
If not, are there any conditions that can be assumed such that this would be true?
For context, I am trying to prove that $$\lim_{n \rightarrow \infty}\left(1 - \lambda\frac{t}{n} - f\left(\frac{t}{n}\right)\right)^n = e^{-\lambda t}$$
where $$\lim_{h \rightarrow 0}\frac{f(h)}{h} = 0$$
On
Notation used are a little bit confusing, I'm assuming that with this
$$\lim_{n \rightarrow \infty}(f^*(n) - o(n))^n = \lim_{n \rightarrow \infty}f^*(n)^n$$
we are assuming $f^*(n)=1 - \lambda\frac{t}{n}$ and $o(n) = f\left(\frac{t}{n}\right)$, and then it is easy to see that
$$\left(1 - \lambda\frac{t}{n} - f\left(\frac{t}{n}\right)\right)^n =\left[\left(1 - \lambda\frac{t}{n} - f\left(\frac{t}{n}\right)\right)^{\frac1{- \lambda\frac{t}{n} - f\left(\frac{t}{n}\right)}}\right]^{t\cdot \frac{- \lambda\frac{t}{n} - f\left(\frac{t}{n}\right)}{\frac t n}} \to e^{-\lambda t}$$
under the condition that $\frac{f\left(\frac t n\right)}{\frac t n} \to 0$.
In the context of your actual goal, you not assign a value to $f(0)$. We can set $f(0) = 0$ without affecting the property that $f(x) = o(x)$ as $x \to 0^+$ (I suspect that might be what you really want, and not $f(x) = o(x)$ as $x \to 0$).
Anyway, your goal is clear if $t = 0$, so let $t \not= 0$. Since you implicitly want $\lambda$ and $t$ to be constant, for $t \not= 0$ let $g(x) = -f(tx)$. Then $g(x) = o(x)$ as $x \to 0^+$, so your task is equivalent to showing $$ \lim_{n \to \infty} (1 - \lambda t/n + g(1/n))^n = e^{-\lambda t}. $$ If $\lambda = 0$, this becomes $\lim_{n \to \infty} (1 + g(1/n))^n = 1$. To verify this, it suffices to take logarithms of both sides (a continuous function) and show $\lim_{n \to \infty} n\log(1 + g(1/n)) = 0$.
As $x \to 0$, $\log(1 + x) = O(x)$, so as $n \to \infty$, $n\log(1 + g(1/n)) = O(ng(1/n))$. We have $ng(1/n) \to 0$ as $n \to \infty$ since $g(x) = o(x)$ as $x \to 0^+$, so $\lim_{n \to \infty} n\log(1 + g(1/n)) = 0$. Now exponentiate both sides of that to see $\lim_{n \to \infty} (1 + g(1/n))^n = 1$.
If $\lambda \not= 0$, then set $u = -\lambda t$ and your limit is equivalent to showing $$ \lim_{n \to \infty} (1 + u/n + g(1/n))^n = e^u $$ when $g(x) = o(x)$ as $x \to 0^+$. Take logarithms of both sides and use he estimate $\log(1+x) = x + O(x^2)$ as $x \to 0$.
Since $$ 1 + \frac{u}{n} + g\left(\frac{1}{n}\right) = 1 + \frac{1}{n}\left(u + \frac{g(1/n)}{1/n}\right), $$ your task can be expressed as $$ \lim_{n \to \infty} \left(1 + \frac{u + \varepsilon(n)}{n}\right)^n = e^u $$ where $u$ is constant and $\varepsilon(n) \to 0$ as $n \to \infty$. The minus signs, $\lambda$, and $t$ in your formulation of the problem are distractions even if they were present in the original setting.