# How do you find the limit of #(x(1-cosx))/tan^3x# as #x->0#?

The limit that is still a problem is the middle limit.

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To find the limit of (x(1-cosx))/tan^3x as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (1-cosx + xsinx)/(3tan^2xsec^2x). Evaluating this expression as x approaches 0, we find that the limit is 1/3.

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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.

- How do you evaluate #[ ( 1 + 3x )^(1/x) ]# as x approaches infinity?
- How do you prove the statement lim as x approaches -1.5 for # ((9-4x^2)/(3+2x))=6# using the epsilon and delta definition?
- How do you find the limit #lim_(h->0)((4+h)^2-16)/h# ?
- How do you evaluate the limit #abs(x+1)+3/x# as x approaches -3?
- What are all horizontal asymptotes of the graph #y=(5+2^x)/(1-2^x)# ?

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