MHB How to Use the Squeeze Theorem to Find This Limit?

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The limit of x^2 • sin(1/x) as x approaches 0 can be determined using the Squeeze Theorem. The sine function is bounded between -1 and 1, leading to the inequalities -x^2 ≤ x^2 • sin(1/x) ≤ x^2. As x approaches 0, both -|x^2| and |x^2| converge to 0. Therefore, by the Squeeze Theorem, the limit of x^2 • sin(1/x) is confirmed to be 0. The conclusion is that lim (x^2 • sin(1/x)) as x approaches 0 equals 0.
nycmathdad
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Use the Squeeze Theorem to find the limit.

lim (x^2 • sin(1/x))
x--> 0

Let me see.

-1 ≤ sin (1/x) ≤ 1

-x^2 ≤ x^2 • sin(1/x) ≤ x^2

-|x^2| ≤ x^2 • sin(1/x) ≤ |x^2|

lim -|x^2| as x tends to 0 = 0.

lim |x^2| as x tends to 0 = 0.
.
By the Squeeze Theorem, x^2 • sin(1/x) was squeezed between the limit of -|x^2| as x tends to 0 and the limit of |x^2| as x tends to 0.

Conclusion:

lim (x^2 • sin(1/x)) = 0
x--> 0

The limit is 0.

Correct?
 
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Yes, that is correct.
 
Country Boy said:
Yes, that is correct.

Very cool. What about the other two Squeeze Theorem Proof threads?
 

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