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.
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 .