# How do you use the Squeeze Theorem to show that #limsinx/x# as x approaches infinity?

Refer to explanation

Hence

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To use the Squeeze Theorem to show that lim(sin(x)/x) as x approaches infinity, we can establish two inequalities. First, we know that -1 ≤ sin(x) ≤ 1 for all x. Therefore, we have -1/x ≤ sin(x)/x ≤ 1/x.

Next, we need to show that the limits of -1/x and 1/x as x approaches infinity are both zero.

For -1/x, as x approaches infinity, the denominator grows infinitely large, causing the fraction to approach zero.

Similarly, for 1/x, as x approaches infinity, the denominator also grows infinitely large, resulting in the fraction approaching zero.

Since -1/x ≤ sin(x)/x ≤ 1/x and both -1/x and 1/x approach zero as x approaches infinity, by the Squeeze Theorem, we can conclude that lim(sin(x)/x) as x approaches infinity is also zero.

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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.

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.

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.

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