How do you find #lim sin(1/x)# as #x->0#?

Answer 1

The limit does not exist.

On any interval containing #0#, #sin(1/x)# takes on the values of #1# and #-1# infinitely many times.
Therefore, the values of #sin(1/x)# cannot be approaching one single value.

graph{sin(1/x) [-2.38, 3.094, -1.216, 1.522]}

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Answer 2

To find the limit of sin(1/x) as x approaches 0, we can use the squeeze theorem. Since the range of the sine function is between -1 and 1, we can conclude that -1 ≤ sin(1/x) ≤ 1 for all x ≠ 0. As x approaches 0, the function oscillates infinitely between -1 and 1. Therefore, the limit of sin(1/x) as x approaches 0 does not exist.

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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.

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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