Prove there is sequence $(t_n)$ such that: i) $t_n →\infty $, ii)$ f'(t_n) =0$, and iii) $ \lim f(t_n) = 1$. Let $f : [0, \infty) \rightarrow \mathbb{R} $ be a bounded, differentiable function satisfying $\underset{ t \to \infty } \lim \ inf \ f(t) = 0$ and $\underset{ t→∞ } \lim \sup \ f(t) = 1$. Prove that there is a sequence $(t_n)^∞ _{n=1}$ such that: i) $t_n →\infty \ as \  n \to \infty$, ii)$ f'(t_n) =0$ for all $n \in N$, and iii) $ \underset{n \to \infty}\lim f(t_n) = 1$.
(i) Since, every bounded sequence in $R^k$ contains a convergent subsequence, there exists a sequence $(t_n)$ such that $t_n \to \infty$ as $n \to \infty$.
(iii) Since $\underset{ t→∞ } \lim \sup \ f(t) = 1$, we have $ \underset{n \to \infty}\lim f(t_n) = 1$.
(ii) To show $ f'(t_n) =0$ , how to use  $\underset{ t \to \infty } \lim \ inf \ f(t) = 0$ and $\underset{ t→∞ } \lim \sup \ f(t) = 1$.
If $f:[0,\infty)\to\mathbb{R}$ is bounded then $|f(x)|&lt;M$ for all $x\in[0,\infty)$. You are given that $\liminf_{t\to\infty}f(t)=0$ and $\limsup_{t\to\infty}f(t)=1$, so $f$ must be bouncing back and forth in some manner, think of oscillations. The assumption that $f$ is differentiable, and hence continuous, is important now because it tells us that as $f$ is oscillating, it does so continuously and so it must have infinitely many turning points.
Can you use this now to find $(t_{n})$?
To get condition (ii), you are going to have to explicitely define the $t_n$ to satisfy this condition, it can't stem from a property about the limits.
Here is a way to do it: first define a sequence $s_n$ such that:

$0&lt;f(s_n)&lt;1$

$s_n$ is increasing

$s_n \to \infty$

$\forall n, \exists x \in ]s_n, s_{n+1}[:\ f(x) &gt; \max\big(f(s_n), f(s_{n+1}),  1 - \frac{1}{n}\big)$


This can easily be constructed by induction: take any $s_1$ such that $0&lt;f(s_1)&lt;1$. Once you have $s_n$, since $f(s_n)&lt;1$ and $\lim \sup f = 1$, there exists $x_0 &gt; s_n+1$ such that $f(x_0) &gt; f(s_n)$ and $f(x_0)&gt;1-\frac{1}{n}$. And then since $\lim \inf f = 0$, you can choose $s_{n+1} &gt; x_0$ such that $f(s_{n+1}) &lt; f(x_0)$ and $f(s_{n+1}) &lt; 1$. Simply note that $s_{n+1} &gt; x_0 &gt; s_n+1$, so $(s_n)$ is increasing and $s_n \to +\infty$. The fourth point is obvious by construction.
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Once we have this sequence, just define $t_n = \mbox{argmax}_{t \in [s_n, s_{n+1}]} f(t)$. Using the fourth point, $t_n \in ]s_n, s_{n+1}[$,and since $f$ is differentiable, $f'(t_n) = 0$. Also, there exists $x_0 \in ]s_n, s_{n+1}[$ such that $f(x_0) &gt; 1 - \frac{1}{n}$ so $f(t_n) \ge f(x_0) &gt; 1 - \frac{1}{n}$. Thus, $f(t_n) \to 1$.
Finally, you have $t_n \ge s_n$, so $t_n \to +\infty$.