Question #e8be5

1 Answer
Nico Ekkart · Tony B
Dec 23, 2015

The limit is #0#

Explanation:

The easiest method to find this is to use the squeeze theorem. This theorem states the following in this case: suppose you find two functions #g(x)# and #h(x)# for which holds that #g(x) <= xsin(1/x) <= h(x)#.
If #lim_{x to 0} g(x) = L = lim_{x to 0} h(x)#, then it definitely holds that:
#lim_{x to 0} xsin(1/x)=L#.

The hardest part is to find two functions for which this holds.
You can do this by noticing that the sine function can only go from #-1# to #1#. This means that the function is always bigger than #-1*|x|# and always smaller than #1*|x|#:
#-|x|<= xsin(1/x)<=|x|#

Let's look at the limits of the two functions:
#lim_{x to 0} -|x| = lim_{x to 0} |x|=0#

This means that #lim_{x to 0} xsin(1/x)=0#

Tony B

Tony B

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