0 times sin(infinity)

If you are asked to evaluate the limit as x tends towards 0 of x[sin(1/x)], you would get 0 times sin(infinity) with quite basic algebraic manipulation. Now of course, we know that sin(infinity) does not exist. My question is, would it be mathematically incoherent to neglect that fact, and nonetheless proceed to say that 0 times anything is zero, regardless of the fact that said anything does not exist? I know that the answer of zero is correct through the use of the squeeze theorem, but my question is about the nature of multiplication through zero.

If you are asked to evaluate the limit as x tends towards 0 of x[sin(1/x)], you would get 0 times sin(infinity) with quite basic algebraic manipulation. Now of course, we know that sin(infinity) does not exist. My question is, would it be mathematically incoherent to neglect that fact, and nonetheless proceed to say that 0 times anything is zero, regardless of the fact that said anything does not exist? I know that the answer of zero is correct through the use of the squeeze theorem, but my question is about the nature of multiplication through zero.

It is indeed incoherent. Let's try applying your argument to a different example. Consider the limit as x tends towards 0 of x(1/x). Subbing in 0, we get 0 times (1/0) = 0 times something = 0. But obviously we know that x(1/x) is constant (when defined) and the limit is 1.

$0\times \sin (\infty)$ is a meaningless piece of notation.

$\lim_{x\to 0} x\sin (1/x)$ has a very specific meaning and is equal to 0

It is indeed incoherent. Let's try applying your argument to a different example. Consider the limit as x tends towards 0 of x(1/x). Subbing in 0, we get 0 times (1/0) = 0 times something = 0. But obviously we know that x(1/x) is constant (when defined) and the limit is 1.

^This


1/x is not bounded, but sin(1/x) is.

Wait, also:

Would a different way to look at it be this?

As x approaches zero, the limit of sin(1/x) remains undefined while the function gets faster, whereas the limit of y=x itself is zero. So instead of solving this algebraically, we can solve it graphically. The function of sin (1/x) is squeezed into a tendency towards 0 by the the function of y=x, thus forcing the limit of x sin(1/x) to be 0. Am I going wrong anywhere?