# http://s14.postimage.org/gmfev9vtd/Lady_Pank.pngPlease solve this problem via l'hopitals rule

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### 1 Answer

You need to remember the formula that gives the area of circular segment, such that:

`A(theta) = (R^2/2)(theta - sin theta)`

You need to evaluate the area `B(theta), ` hence, you need to find the length of chord PR and the height h, such that:

`PR = 2R sin(theta/2)`

`h = R - R(1 - cos(theta/2)) = Rcos(theta/2)`

Hence, evaluating the area `B(theta)` yields:

`B(theta) = PR*h/2 => B(theta) = 2R sin(theta/2)Rcos(theta/2)/2`

`B(theta) = R^2*sin(2*theta/2)/2 => B(theta) = (R^2 sin theta)/2`

You need to evaluate the following limit, such that:

`lim_(theta -> 0) (A(theta))/(B(theta)) = lim_(theta -> 0) ((R^2/2)(theta - sin theta))/((R^2 sin theta)/2)`

Reducing duplicate factors yields:

`lim_(theta -> 0) (A(theta))/(B(theta)) = lim_(theta -> 0) (theta - sin theta)/(sin theta)`

Substituting 0 for theta yields:

`lim_(theta -> 0) (theta - sin theta)/(sin theta) = (0 - sin 0)/ (sin 0)`

`lim_(theta -> 0) (theta - sin theta)/(sin theta) = 0/0`

You may use l'Hospital's theorem such that:

`lim_(theta -> 0) (theta - sin theta)/(sin theta) = lim_(theta -> 0) ((theta - sin theta)')/((sin theta)')`

`lim_(theta -> 0) ((theta - sin theta)')/((sin theta)') = lim_(theta -> 0) (1 - cos theta)/(cos theta)`

`lim_(theta -> 0) (1 - cos theta)/(cos theta) = (1 - cos 0)/(cos 0)`

`lim_(theta -> 0) (1 - cos theta)/(cos theta) = (1 - 1)/1`

`lim_(theta -> 0) (1 - cos theta)/((cos theta)) = 0/1 = 0 `

**Hence, evaluating the given limit `lim_(theta -> 0) (A(theta))/(B(theta))` using l'Hospital's theorem `lim_(theta -> 0) (A(theta))/(B(theta)) = 0.` **