How do you graph #y = x sin(1/x)#?

1 Answer
Mar 29, 2016

As #sin(theta) in [-1,1]#, the #x# prior to #sin(1/x)# acts as a scaling factor. As #x# grows large, the amplitude of the oscillations of the sine function also grow. Similarly, as #x# approaches #0#, the amplitude shrinks.

Next, looking at #sin(1/x)# we note that #1/x->oo# as #x->0#. This means that as #x->0# the sine function cycles through periods of #2pi# more and more rapidly. Similarly, #1/x->0# as #x->+-oo#, meaning it will take greater and greater changes in #x# to go through a full period of #2pi#. After #|x| > 1/(pi)# there will be no further oscillations, as we will have #|1/x| in (0,pi)#.

The resulting graph will have oscillations which grow in amplitude and are stretched further apart as #x# is further from #0#, soon ceasing to oscillate at all, and which have smaller amplitudes and wilder oscillations closer to #0#.

graph{xsin(1/x) [-1.5, 1.5, -0.75, 0.75]}