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It would seem at first glance that for any $x \neq 0$ we can cancel out the $x^3$ factors in numerator and denominator and have $f(x) = x$.

MITOCW problem set 1 says there is a removable discontinuity at $x=0$ (problem 1D-3c, and the official solutions)

What is in fact the difference between $f(x) = x$ and $f(x) = x^4/x^3$ in terms of continuity?

For example, if I start with $f(x) = x$ and then multiply by $x^3/x^3$ is this different from starting from $f(x) = x^4/x^3$, with regards to what happens at $x=0$?

xoux
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  • $x^4/x^3$ is not defined at the origin. Away from the origin, the functions are identical. The limit for both at the origin is $0$ and does not depend on the value at $0$. The function $x^4/x^3$ is not continuous at the origin, because its value there is undefined. See here for a great explanation: https://math.stackexchange.com/a/1822706/754927 – Ben Apr 18 '21 at 02:04
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    The difference is that $f(x) = x$ is the function that maps to $x$ no matter what $x$ is. While $g(x) = \frac {x^4}{x^3}$ is a function that maps to $x$ if $x\ne 0$ but is completely undefined and does not exist if $x = 0$. $f$ is continuous everywhere. $g$ is continuous everywhere except at $x =0$ where it is not defined at all. – fleablood Apr 18 '21 at 03:55
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    "if I start with f(x)=x and then multiply by x3/x3 is this different from starting from f(x)=x4/x3, with regards to what happens at x=0?" You can't multiply by $\frac 00$ if $x = 0$ because $\frac 00$ doesn't exist and is not anything at all. – fleablood Apr 18 '21 at 03:57

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