How do you use the Squeeze Theorem to find #lim xcos(1/x)# as x approaches zero?

Answer 1

Use the fact that the cosine function is always between #-1# and #1#, implying that the given function is always between #-|x|# and #|x|#, which both go to zero as #x# goes to zero.

Let #f(x)=x cos(1/x)#, #g(x)=-|x|#, and #h(x)=|x|#. Since #-1 leq cos(1/x) leq 1# for all #x !=0#, it follows that #g(x) leq f(x) leq h(x)# for all #x !=0#.
But #lim_{x->0}g(x)=lim_{x->0}h(x)=0#. Therefore, the Squeeze Theorem can be use to conclude that #lim_{x->0}f(x)=lim_{x->0}x cos(1/x)=0#.
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Answer 2

To use the Squeeze Theorem to find lim xcos(1/x) as x approaches zero, we can establish two functions that "squeeze" the given function.

First, we consider the function f(x) = -1, which is always less than or equal to xcos(1/x) for all x.

Next, we consider the function g(x) = 1, which is always greater than or equal to xcos(1/x) for all x.

Since f(x) ≤ xcos(1/x) ≤ g(x) for all x, and both f(x) and g(x) approach zero as x approaches zero, we can conclude that lim xcos(1/x) as x approaches zero is also zero.

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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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