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How do you find the limit of # (1-cos(x))/x# as x approaches 0?

Answer 1

When in doubt, use L'Hopitals rule, maybe even more then once, especially if there are polynomials on the bottom or constants to be rid of, make them all go poof!

L'Hopitals rule says that: #lim_{x to a} {f(x)}/{g(x)}=lim_{x to a} {f'(x)}/{g'(x)}#

We trying to solve, #lim_{x to 0} {1-cos(x)}/{x}#.

Take the derivative of the top and the bottom functions: #d/{dx} (1- cos (x)) =d/{dx} (1 ) - d/{dx} ( cos (x))=0- (-sin(x))=sin(x)# and #d/{dx} (x)=1#. So, #lim_{x to 0} {1-cos(x)}/{x} =lim_{x to 0} {sin(x)}/{1}=lim_{x to 0} sin(x)# Since #sin(x)# is well behaved at zero, this is just #sin(0)# which is #0#.

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Answer 2

Ezekiel Alexander

To find the limit of (1-cos(x))/x as x approaches 0, we can use L'Hôpital's Rule. By differentiating the numerator and denominator separately, we get sin(x)/1. Evaluating this expression as x approaches 0, we find that the limit is equal to 1.

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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.