Q:

Prove that $\varphi^n(t) \rightarrow 0$ when $n \rightarrow \infty$

I have to prove a lemma:

If $\varphi: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ is monotone increasing and $\varphi(t) < t, \ \forall t \in \mathbb{R}_+$, then $\varphi^n(t) \rightarrow 0$, $(n \rightarrow \infty)$.

To be clear, $\varphi^2(t) = \varphi(\varphi(t))$.
It's just not clear to me how to proceed.
Edit:
as stated by Karolis Juodelė, for the lemma to be true, $\varphi$ must be continuous.

A:

It's not true. Take
$$\varphi(t) = \begin{cases} \frac t 2 & t \leq 1 \\ \frac 1 2 + \frac t 2 & t > 1 \end{cases}$$
Note that although $\lim_{t \to 1} \varphi(t) = 1$ but $\varphi(1) = \frac 1 2$  and $\varphi(t) < t$ holds. Note also that $\lim \varphi^n(t) = 1$ when $t > 1$. For it to be $0$, $\varphi$ should be continuous.
Assume that $\varphi$ in fact is continuous. Note that the sequence $x_n = \varphi(x_{n-1}) = \varphi^n(t)$ is decreasing and bounded, thus converges to $x$. Now, because $\varphi$ is continuous, $\lim x_n = \lim \varphi(x_{n-1}) = \varphi(\lim x_{n-1}) = \varphi (x) = x$. By definition of $\varphi$, this implies $x = 0$

