How do you use the Squeeze Theorem to find #limsin(1/x)# as x approaches zero?
The limit
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To use the Squeeze Theorem to find lim sin(1/x) as x approaches zero, we can establish two functions that "squeeze" the given function.
First, we note that 1 ≤ sin(1/x) ≤ 1 for all x ≠ 0.
Next, we consider two additional functions:

The constant function f(x) = 1, which is always greater than or equal to sin(1/x) for all x ≠ 0.

The constant function g(x) = 1, which is always less than or equal to sin(1/x) for all x ≠ 0.
Since f(x) ≥ sin(1/x) ≥ g(x) for all x ≠ 0, and both f(x) and g(x) approach 1 as x approaches zero, we can conclude that the limit of sin(1/x) as x approaches zero is also 1.
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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