# How do you use the Squeeze Theorem to find #lim sqrt(x)sin(1/x) # as x approaches infinity?

See the explanation section.

If we have been through the proof of

(Or perhaps it was pointed out in our study of trigonometry. See note, below.)

So, for large positive

Observe that

Therefore,

Note

Here is a picture reminder for

For the angle measuring

while

Because the perpendicular distance is the shortest, we believe that, for small positive

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To use the Squeeze Theorem to find the limit of sqrt(x)sin(1/x) as x approaches infinity, we need to find two functions that "squeeze" the given function and have the same limit as x approaches infinity.

First, we can observe that -1 ≤ sin(1/x) ≤ 1 for all x. Therefore, we have -sqrt(x) ≤ sqrt(x)sin(1/x) ≤ sqrt(x) for all x.

Next, we can find the limits of the two bounding functions as x approaches infinity.

The limit of -sqrt(x) as x approaches infinity is -∞.

The limit of sqrt(x) as x approaches infinity is ∞.

Since -∞ ≤ sqrt(x)sin(1/x) ≤ ∞ for all x, and the limits of the bounding functions are -∞ and ∞, respectively, we can conclude that the limit of sqrt(x)sin(1/x) as x approaches infinity is also ∞.

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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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