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The following is taken from some lecture notes on Monte Carlo methods: enter image description here

I have never seen in anywhere else that we can treat $0/0=1$. Why it is okay to do it here?

Sam
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    It's often done (though phrased differently) in cases where it's asymptotically true near 0; for example, when filling in sin(x)/x at x=0. – Alex K Jan 14 '23 at 05:15
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    There is no standard convention of setting $0/0=1$ but any step that has $0/0$ is also multiplied by $0$, so it doesn't matter what value you assign to $0/0$. My interpretation is just that: "we could do case-work or restrict the domains, but if you'll let me define $0/0$ then I can write it like this" – Brian Moehring Jan 14 '23 at 05:31
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    I can't tell if it is the case here, but because it's inside an integral, if $f$ only vanishes on a set of measure zero, then we can basically ignore what happens there, and not affect the value of the integral. A more careful treatment would break the integrals into two regions, one where $f$ isn't zero, and one where it is but it's over a zero measure set. My guess is that after you do that three or four times you see that you get the same thing as if you had done the lazy method. – JonathanZ Jan 14 '23 at 05:32
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    "$0/0=1$" is not a valid statement. It only makes sense in the sense of limits and is even then only correct in the situations where the limit is actually $1$ for $x\to 0$. – Peter Jan 14 '23 at 08:48

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