A function $ f:\mathbb{R}_{\ge 0}\longrightarrow\mathbb{R} $ has the property that $ \lim_{x\to\infty } \frac{1}{x^2}\int_0^x f(t)dt=1. $

[b]a)[/b] Give an example of what $ f $ could be if it's continuous and $ f/\text{id.} $ doesn't have a limit at $ \infty . $
[b]b)[/b] Prove that if $ f $ is nondecreasing then $ f/\text{id.} $ has a limit at $ \infty , $ and determine it.
### Part (a)
We need to find a continuous function \( f \) such that \( \lim_{x \to \infty} \frac{1}{x^2} \int_0^x f(t) \, dt = 1 \) and \( \frac{f(x)}{x} \) does not have a limit as \( x \to \infty \).

1. Consider the base function \( f(x) = 2x \). For this function:
   \[
   \int_0^x f(t) \, dt = \int_0^x 2t \, dt = x^2.
   \]
   Therefore,
   \[
   \frac{1}{x^2} \int_0^x f(t) \, dt = \frac{1}{x^2} \cdot x^2 = 1.
   \]
   This satisfies the given limit condition.

2. To ensure \( \frac{f(x)}{x} \) does not have a limit as \( x \to \infty \), we can modify \( f(x) \) by adding sharp peaks at specific points. For example, let:
   \[
   f(x) = 2x + \sum_{n=1}^{\infty} \chi_{[n, n+1)}(x) \cdot h_n,
   \]
   where \( \chi_{[n, n+1)}(x) \) is the characteristic function of the interval \([n, n+1)\) and \( h_n \) are chosen such that the area under each peak is \( 2^{-n} \).

3. The integral of \( f(x) \) over \([0, x]\) will still be dominated by the \( 2x \) term, ensuring the limit condition is met. However, the peaks will cause \( \frac{f(x)}{x} \) to oscillate and not converge to a single value.

### Part (b)
We need to prove that if \( f \) is nondecreasing, then \( \frac{f(x)}{x} \) has a limit as \( x \to \infty \), and determine this limit.

1. Assume \( f \) is nondecreasing. Given:
   \[
   \lim_{x \to \infty} \frac{1}{x^2} \int_0^x f(t) \, dt = 1,
   \]
   we need to show \( \lim_{x \to \infty} \frac{f(x)}{x} \) exists.

2. Define \( F(x) = \int_0^x f(t) \, dt \). The given condition can be rewritten as:
   \[
   \lim_{x \to \infty} \frac{F(x)}{x^2} = 1.
   \]

3. By the Fundamental Theorem of Calculus, \( F'(x) = f(x) \). Using L'Hôpital's Rule:
   \[
   \lim_{x \to \infty} \frac{F(x)}{x^2} = \lim_{x \to \infty} \frac{f(x)}{2x}.
   \]

4. Since \( \lim_{x \to \infty} \frac{F(x)}{x^2} = 1 \), we have:
   \[
   \lim_{x \to \infty} \frac{f(x)}{2x} = 1 \implies \lim_{x \to \infty} \frac{f(x)}{x} = 2.
   \]

Thus, if \( f \) is nondecreasing, \( \frac{f(x)}{x} \) has a limit at \( \infty \), and this limit is \( 2 \).

\(\blacksquare\)

The final answer is \( \boxed{ 2 } \).