Question
e. Assuming the limit exists, use the same ideas to determine the value of \( a+\frac{b}{\left[a+\frac{b}{a+\frac{b}{a+\cdots}}\right]} \), where \( a \) and \( b \) are positive real numbers. Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. The limit is \( \square \). (Type an exact answer, using radicals as needed.) B. The limit does not exist.
Ask by Chang Hampton. in the United States
Mar 15,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The limit is \( \frac{a + \sqrt{a^2 + 4b}}{2} \).
Solution
Let \( x \) be the value of the continued fraction. Then we have
\[
x = a + \frac{b}{x}.
\]
We can multiply both sides by \( x \) to eliminate the fraction:
\[
x^2 = ax + b.
\]
Rearrange this into a quadratic equation:
\[
x^2 - ax - b = 0.
\]
Since \( x \) represents a length (or a positive number, given that \( a \) and \( b \) are positive), we take the positive solution of the quadratic formula:
\[
x = \frac{a \pm \sqrt{a^2 + 4b}}{2}.
\]
Thus, the limit is
\[
\boxed{\frac{a+\sqrt{a^2+4b}}{2}}.
\]
The correct choice is A.
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Simplify this solution Extra Insights
To solve for the value of \( L = a + \frac{b}{L} \), we can start by recognizing that this expression leads to a quadratic equation. Multiplying both sides by \( L \) gives us: \[ L^2 = aL + b \] Rearranging this into standard quadratic form: \[ L^2 - aL - b = 0 \] We can apply the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ L = \frac{a \pm \sqrt{a^2 + 4b}}{2} \] Since \( a \) and \( b \) are positive, we only take the positive root (to maintain \( L \) as a positive real number). Thus, we find: \[ L = \frac{a + \sqrt{a^2 + 4b}}{2} \] A. The limit is \( \frac{a + \sqrt{a^2 + 4b}}{2} \).
