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It is well known that the function $f(x)=1/x$ is not defined for $x=0$. However, simply multiplying $f$ by the function $g(x)=x$ gives a constant, very well defined, function, even at $x=0$. How can it be that multiplying by $g(0)=0$ something which doesnt exists ($f(0)$) can actually yield a result ?

Moreover, if I multiply the whole $f(x)$ by $0$ it obviously gives me $0$ at $x=0$ where i had $1$ in the previous case. This seems to indicate that either :

  • there is something more to the function $f$ at $0$ than not being defined
  • there is something wrong with the way I think about the product of two functions.
user280229
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    Remember, the domain of the product $(f \cdot g)(x) = f(x) \cdot g(x)$ is the intersection of the domains of $f$ and $g$; the product is only defined at $x$-values where both functions are defined. – pjs36 Oct 14 '15 at 13:12
  • I think a point of confusion is that it seems you want to say $f(x)g(x)$ precisely equals $h(x)$ where $h(x) = 1$. This is only true everywhere except zero. – kmeis Oct 14 '15 at 13:12
  • The function $a(x)=\frac xx$ is not equal to the function $b(x)=1$. – Aditya Agarwal Oct 14 '15 at 13:39

1 Answers1

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The answer lies in the fact that:
Let $a(x)=\frac xx$. Now if we cancel the $x$(s), we are assuming that $x\neq0$. So our new function becomes, $a(x)=1;x\neq0$. But $b(x)=1$ is also defined for $0$. So $b(x)$ is continuous at $x=0$ but $a(x)$ is not. Two functions are said to be "equal" when they have same domain and they yield same value at ALL values of the domain .

Thus, $a(x)\neq b(x).$

manifold
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Aditya Agarwal
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