How do you use the Squeeze Theorem to find #lim x^2 (Sin 1/x)^2 # as x approaches zero?

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Answer 1 · 2015-10-11T17:12:33

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

Assuming you meant #x^2sin^2(1/x)#

We know that

#0 <= sin^2(theta) <= 1#

So for #theta = 1/x# we have

#0 <= sin^2(1/x) <= 1#

Multiplying both sides by #x^2#

#0 <= x^2sin^2(1/x) <= x^2#

Since

#lim_(x rarr 0)0 = lim_(x rarr 0) x^2 = 0#

The squeeze theorem tells us that

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