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Truncation error is introduced by replacing an exact operation with a finite approximation. It remains even when all arithmetic is exact. For $f(x)=x^2$, the forward finite difference for its derivative at $x=1$ givesThus the truncation error is exactly $h$: $h=0.1$ gives $2.1$ and $h=0.01$ gives $2.01$. In finite precision, however, the numerator subtracts nearly equal values, and their roundoff error can be amplified by division by $h$. For smooth data with bounded evaluation errors of scale $u$, a typical bound has competing terms $C_{1}h+C_{2}u/h$. Smaller $h$ reduces the first term and increases the second; with fixed positive constants the balancing scale is $h\sim \sqrt{u}$. This is why blindly refining a finite difference eventually stops improving it.