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

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

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