Question

Question #51a7e

Answer 1

No the limiti is `0`, because when `xrarroo`, `1/xrarr0` and so `sin0=0`.

These are limits they don't exist:

`lim_(xrarr+oo)sinx`

or

`lim_(xrarr0)sin(1/x)`.

(`sinoo` does not exist).

Answer 2

If someone told you the limit does not exist for that reason, they've probably confused this question

`lim_(xrarroo) sin (1/x)` which is `0`

With this one

`lim_(xrarr0) sin (1/x)` which down not exist because the values cover `[-1, 1]` over shorter and shorter intervals as `xrarr0`

Answer 3

Actually, that WOULD be correct if you were finding the limit of `sin(x)`. As `x` approaches infinity, `sin(1/x)` just becomes `sin(0)`, which is `0`. graph{sin(1/x) [-9.775, 10.225, -4.78, 5.22]}