Is the Error O(1/n) in Linear Interpolation Correct?

[AxiomOfChoice]

Consider the following situation. You know, for a given function $f$ and $i,k \in \mathbb N \cup \{ 0 \}$, $n\in \mathbb N$, that

$$f \left( \frac kn \right) = \frac{i}{\sqrt n}.$$

In addition, you know that either

$$f \left( \frac{k+1}{n} \right) = \frac{i-1}{\sqrt n}$$

-OR-

$$f \left( \frac{k+1}{n} \right) = \frac{i+1}{\sqrt n}.$$

And suppose that you linearly interpolate between k/n and (k+1)/n. The book I'm reading claims that, for any t between k/n and (k+1)/n, we have

$$f(t) - f \left( \frac{\lfloor tn \rfloor}{n} \right) = O(1/n).$$

where of course I have referred to the floor function. (Incidentally, if $\frac kn \leq t \leq \frac{k+1}{n}$ as assumed, doesn't it follow that $\lfloor nt \rfloor = k$?) Question: IS THIS (the part about O(1/n)) RIGHT? I don't think it is.

 

[AxiomOfChoice]

I've calculated, and it seems to me that if the FIRST option obtains, we have

$$f(t) - f \left( \frac{\lfloor tn \rfloor}{n} \right) = \frac{k - tn}{\sqrt n},$$

whereas if the SECOND option obtains, we have

$$f(t) - f \left( \frac{\lfloor tn \rfloor}{n} \right) = \frac{tn - k}{\sqrt n}.$$

So I know EXACTLY what the errors are as a function of $t$, and in either case they are bounded by $1/\sqrt n$, simply because $|tn - k| \leq 1$. Have I done something wrong?