Given the piecewise function: #f(x)= x^2sin(1/x), x != 0# and #0, x = 0# Is#f# continuous at #x=0# and if so, prove it?

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Answer 1 · 2017-03-17T15:14:04

Yes. By the definition of #f# we have #f(0) = 0#, and by the squeeze theorem we have #lim_(xrarr0)f(x) = 0#