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

Answer 1

Emily Ammerman

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

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

We are aware of 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#

As per the squeeze theorem,

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

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Answer 2

Scarlett Allison

To use the Squeeze Theorem to find the limit of the function lim x^2 (Sin 1/x)^2 as x approaches zero, we need to find two other functions that "squeeze" the given function and have the same limit as x approaches zero.

First, we can observe that -1 ≤ Sin(1/x) ≤ 1 for all x ≠ 0. Therefore, we can square this inequality to get 0 ≤ (Sin(1/x))^2 ≤ 1.

Next, we multiply the inequality by x^2 to get 0 ≤ x^2(Sin(1/x))^2 ≤ x^2.

Since the limit of x^2 as x approaches zero is also zero, we can conclude that the limit of x^2(Sin(1/x))^2 as x approaches zero is also zero, based on the Squeeze Theorem.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.