Wolfram|Alpha states that the infinite continued fraction $$\cfrac{1}{0+\cfrac{1}{0+\cdots}}=0.$$ Assuming $[0;0,0,\ldots]$ exists implies that the continued fraction is $1$, since $x=\dfrac{1}{0+\cfrac{1}{0+\cdots}}=\dfrac{1}{x}$ implies $x=1$ (ignoring the negative value). This, along with the divide by zero error, suggests W|A is wrong.
Is this an error on W|A's part? If not, is this just a convention, and is there a similar convention for $$\cfrac{0}{0+\cfrac{0}{0+\cdots}}?$$
As Gerry indicates in his comment: the definition of an infinite continued fraction is the limit of its finite truncations. Here none of the truncations are well-defined, so certainly neither is the limit.