For $k=1,2,\dots$, let $f_k$ be the number of times
\[\sin\left(\frac{k\pi x}{2}\right)\]
attains its maximum value on the interval $x\in[0,1]$. Compute
\[\lim_{k\rightarrow\infty}\frac{f_k}{k}.\]
1. We start by analyzing the function \(\sin\left(\frac{k\pi x}{2}\right)\). The sine function attains its maximum value of 1 when its argument is an odd multiple of \(\frac{\pi}{2}\). Therefore, we need:
   \[
   \frac{k\pi x}{2} = \frac{\pi}{2} + n\pi \quad \text{for some integer } n.
   \]
   Simplifying, we get:
   \[
   kx = 1 + 2n \quad \text{or} \quad kx = 1 + 2n \quad \text{mod} \quad 4.
   \]

2. To find the values of \(x\) that satisfy this condition within the interval \([0, 1]\), we solve for \(x\):
   \[
   x = \frac{1 + 2n}{k}.
   \]
   For \(x\) to be in the interval \([0, 1]\), we need:
   \[
   0 \leq \frac{1 + 2n}{k} \leq 1.
   \]
   This implies:
   \[
   0 \leq 1 + 2n \leq k.
   \]
   Simplifying, we get:
   \[
   -1 \leq 2n \leq k - 1.
   \]
   Dividing by 2:
   \[
   -\frac{1}{2} \leq n \leq \frac{k - 1}{2}.
   \]
   Since \(n\) must be an integer, the range of \(n\) is:
   \[
   0 \leq n \leq \left\lfloor \frac{k - 1}{2} \right\rfloor.
   \]

3. The number of integers \(n\) in this range is:
   \[
   \left\lfloor \frac{k - 1}{2} \right\rfloor + 1.
   \]
   As \(k\) approaches infinity, the term \(\left\lfloor \frac{k - 1}{2} \right\rfloor\) approaches \(\frac{k}{2}\). Therefore, the number of such \(n\) is approximately:
   \[
   \frac{k}{2}.
   \]

4. However, we need to count the number of times \(\sin\left(\frac{k\pi x}{2}\right)\) attains its maximum value, which occurs when \(kx \equiv 1 \pmod{4}\). This means we need to count the number of integers \(k\) such that:
   \[
   kx \equiv 1 \pmod{4}.
   \]
   For large \(k\), the number of such \(x\) values is approximately \(\frac{k}{4}\).

5. Therefore, the limit we are interested in is:
   \[
   \lim_{k \to \infty} \frac{f_k}{k} = \frac{1}{4}.
   \]

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{4} } \).