How do you find the limit of # x^2 sin(1/x)# as x approaches 0?

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Answer 1 · 2016-08-30T17:19:21

Use the squeeze theorem.

We know from trigonometry that

#-1 <= sin(1/x) <- 1# for all #x != 0#.

Important: for #lim_(xrarr0)# we don't care what happens when #x = 0#.

Since #x<2 > 0# for all #x != 0#, we can multiply through by #x^2# to get

#-x^2 = x^2 sin(1/x) <= x^2#.

Clearly #lim_(xrarr0) (-x^2) = 0# and #lim_(xrarr0) x^2 = 0#, so, by the squeeze theorem,

#lim_(xrarr0) x^2 sin(1/x) = 0#.