# How do you find the limit of #x^2cos(pi / x)# as x approaches 0?

cos(1/x) oscillates between 1 and -1 really fast when aproaching to 0 but it always is a finite number. If you are multipliying this oscillating function to x^2 which modullates the oscillation it will oscillate between x^2 and -x^2. So if you make the limit going to 0 it will "oscillating" between +0 and -0 which you see is 0.

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To show this, use the squeeze (pinch, sandwich) theorem.

Note that

So the squeeze theorem (or whatever you call it) tells us that

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To find the limit of x^2cos(pi / x) as x approaches 0, we can use the squeeze theorem.

First, we note that -1 ≤ cos(pi / x) ≤ 1 for all x.

Next, we multiply both sides of the inequality by x^2 (since x^2 is always non-negative) to get:

-x^2 ≤ x^2cos(pi / x) ≤ x^2

As x approaches 0, both -x^2 and x^2 approach 0.

Therefore, by the squeeze theorem, the limit of x^2cos(pi / x) as x approaches 0 is 0.

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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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