What is the limit of #xsinx# as x approaches infinity?

1 Answer
Jul 18, 2016

The limit does not exist. See below.

Explanation:

We can determine the result by pure intuition.

We know that #sinx# alternates between #-1# and #1#, from negative infinity to infinity. We also know that #x# increases from negative infinity to infinity. What we have, then, at large values of #x# is a large number (#x#) multiplied by a number between #-1# and #1# (due to #sinx#).

This means the limit does not exist. We do not know if #x# is being multiplied by #-1# or #1# at #oo#, because there is no way for us to determine that. The function will essentially alternate between infinity and negative infinity at large values of #x#. If, for example, #x# is a very large number and #sinx=1#, then the limit is infinity (large positive number #x# times #1#); but #(3pi)/2# radians later, #sinx=-1# and the limit is negative infinity (large positive number #x# times #-1#).