The following expression, where the process continues indefinitely, is called a continued fraction. Complete parts (a) through (e). \[ 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}} \] c. Using computation and/or graphing, estimate the limit of the sequence, if it exists. Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. The limit appears to be approximately \( \square \) . (Type an integer or a decimal rounded to three decimal places as needed.) B. The limit does not appear to exist.

We begin by letting \[ x = 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}}. \] Because the process continues indefinitely, the expression inside the denominator is also \( x \). This gives \[ x = 2 + \frac{1}{x}. \] Multiplying both sides of the equation by \( x \) to clear the fraction, we have \[ x^2 = 2x + 1. \] Rearrange the equation to form a quadratic equation: \[ x^2 - 2x - 1 = 0. \] We solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}, \] where \( a = 1 \), \( b = -2 \), and \( c = -1 \). Substituting these values in, we obtain \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2}. \] Since \( \sqrt{8} = 2\sqrt{2} \), this simplifies to \[ x = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}. \] We discard the negative solution because \( x \) must be positive (as it is defined by a continued fraction composed of positive numbers). Therefore, \[ x = 1+\sqrt{2}. \] Approximating \( \sqrt{2} \) as about \( 1.414 \), we have \[ 1+\sqrt{2} \approx 1 + 1.414 = 2.414. \] Thus, the limit of the sequence is approximately \[ 2.414. \]