Question

What are the asymptote(s) and hole(s), if any, of `f(x) = xsin(1/x)`?

Answer 1

Refer below.

Explanation:

Well, there is obviously a hole at `x=0`, since division by `0` is not possible.

We can graph the function:
graph{xsin(1/x) [-10, 10, -5, 5]}

There are no other asymptotes or holes.

Answer 2

`f(x)` has a hole (removable discontinuity) at `x=0`.

It also has a horizontal asymptote `y=1`.

It has no vertical or slant asymptotes.

Explanation:

Given:

`f(x) = x sin(1/x)`

I will use a few of properties of `sin (t)`, namely:

• `abs(sin t) <= 1" "` for all real values of `t`.

• `lim_(t->0) sin(t)/t = 1`

• `sin(-t) = -sin(t)" "` for all values of `t`.

First note that `f(x)` is an even function:

`f(-x) = (-x) sin(1/(-x)) = (-x)(-sin(1/x)) = x sin(1/x) = f(x)`

We find:

`abs(x sin(1/x)) = abs(x) abs(sin (1/x)) <= abs(x)`

So:

`0 <= lim_(x->0+) abs(x sin(1/x)) <= lim_(x->0+) abs(x) = 0`

Since this is `0`, so is `lim_(x->0+) x sin(1/x)`

Also, since `f(x)` is even:

`lim_(x->0^-) x sin(1/x) = lim_(x->0^+) x sin(1/x) = 0`

Note that `f(0)` is undefined, since it involves division by `0`, but both left and right limits exist and agree at `x=0`, so it has a hole (removable discontinuity) there.

We also find:

`lim_(x->oo) x sin(1/x) = lim_(t->0^+) sin(t)/t = 1`

Similarly:

`lim_(x->-oo) x sin(1/x) = lim_(t->0^-) sin(t)/t = 1`

So `f(x)` has a horizontal asymptote `y=1`

graph{x sin(1/x) [-2.5, 2.5, -1.25, 1.25]}