# How do you use the Squeeze Theorem to evaluate #lim x->0# of #x^2cos(1/(42x))#?

Observing that

Using the squeeze theorem, we have

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To evaluate ( \lim_{x \to 0} x^2 \cos\left(\frac{1}{42x}\right) ) using the Squeeze Theorem, we can find two functions that "squeeze" the given function as ( x ) approaches 0. Since the cosine function is bounded between -1 and 1, we can use these bounds to find two functions.

Let's consider the function ( g(x) = x^2 ) and ( h(x) = -x^2 ). Then, ( -x^2 \leq x^2 \cos\left(\frac{1}{42x}\right) \leq x^2 ) for all ( x ) except when ( x = 0 ).

Now, taking the limit of ( g(x) ) and ( h(x) ) as ( x ) approaches 0: [ \lim_{x \to 0} g(x) = \lim_{x \to 0} x^2 = 0 ] [ \lim_{x \to 0} h(x) = \lim_{x \to 0} -x^2 = 0 ]

Since ( g(x) ) and ( h(x) ) both approach 0 as ( x ) approaches 0, and ( g(x) \leq x^2 \cos\left(\frac{1}{42x}\right) \leq h(x) ) for all ( x ) near 0, by the Squeeze Theorem, ( \lim_{x \to 0} x^2 \cos\left(\frac{1}{42x}\right) = 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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