How do you evaluate the continued fraction #f(e)=e-1+1/(e-1+1/(e-1+1/(e-1+1/(e-1+1/...# ? Algebra Linear Equations Two-Step Equations and Properties of Equality 1 Answer Shwetank Mauria · A. S. Adikesavan · Cesareo R. Jul 8, 2016 #(e-1+sqrt(e^2-2e+5))/2=2.17752#, nearly. Explanation: Let #f(e)=x#, then #x=e-1+1/x#. Hence, #x^2=ex-x+1# or #x^2-x(e-1)-1=0# and using quadratic formula #x=(e-1+-sqrt(e^2-2e+5))/2# #x>e-1# So, #x=(e-1+sqrt(e^2-2e+5))/2=2.17752, #nearly. Answer link Related questions How do you solve two step equations? How do you check solutions to two step equations? What is an example of a two step equation with no solution? How do I check to see if the solution is 1 for the equation #2x+1=3#? Is there more than one way to solve a 2 step equation? How do you solve #-m+3=3#? How do you solve #-5y-9=74#? How do you solve #5q - 7 = \frac{2}{3}#? How do you solve #0.1y + 11 =0#? How do you solve #\frac{5q-7}{12} = \frac{2}{3}#? See all questions in Two-Step Equations and Properties of Equality Impact of this question 1641 views around the world You can reuse this answer Creative Commons License