Question

What is the limit as x approaches 0 of `sin(1/x)`?

Answer 1

This limit does not exist, or with other words, it diverges.

You can see this by substituting `u=1/x`. Then, as `x` approaches zero, `u` approaches infinity.
Therefore:
`lim_(x->0)sin(1/x) = lim_(u->oo)sin(u)`
This limit does not exist, for the sine is a periodic fluctuating function.