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

2 Answers
Feb 4, 2018

Answer:

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.

Feb 4, 2018

Answer:

#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]}