An algorithm for approximating the sqrt (square root) ofAis the following: To approximate sqrtA, begin with the initial approximation (obtained by "eyeballing it") ofx0. Then, let the next approximation bex1 = ((x0)^2 +A)/(2*x0), thenx2= ((x1)^2 +A)/(2*x1) and generally, use the recursive definitionxn+1 = ((xn)^2 +A)/(2*xn).

Assuming that the sequence of successive approximations converges to a limitL, show thatL= sqrtA.

I think that I need to take limits of both sides ofxn+1 = ((xn)^2 +A)/(2*xn). I do not know how I would go about doing this.