Why check for !isNaN() after isFinite()?

Question

I came across the goog.math.isFiniteNumber function in the Google Closure Library. What it does is checking whether a given number is both finite and not NaN<

Answer 1

If isFinite worked the way isFiniteNumber did, then there would be no reason to write isFiniteNumber. There's probably some browser out there somewhere that treats NaN as finite.