Q.11 Let a = √(5+1)/2 and b = √(5-1)/2. Then, limn→∞ (an + bn)/(an - bn) (A) is 1 (B) is ½ (C) is 0 (D) does not exist

Q.11 Let a = √(5+1)/2 and b = √(5-1)/2. Then, limn→∞ (an + bn)/(an – bn)

  • (A) is 1
  • (B) is ½
  • (C) is 0
  • (D) does not exist

Limit of (an + bn)/(an − bn) as n tends to infinity

The limit \(\lim_{n\to\infty}\frac{a^n+b^n}{a^n-b^n}\) for \(a=\frac{\sqrt5+1}{2}\) and \(b=\frac{\sqrt5-1}{2}\) equals 1, so the correct option is (A). [web:3]

Introduction

This article explains how to find the limit of (a^n + b^n)/(a^n − b^n) as n tends to infinity when \(a=\frac{\sqrt5+1}{2}\) and \(b=\frac{\sqrt5-1}{2}\). [web:3]

The solution uses properties of exponential growth and shows why the correct multiple‑choice option is 1, while the other options fail. [web:3]

Step‑by‑step solution

Let \(a=\frac{\sqrt5+1}{2}\) and \(b=\frac{\sqrt5-1}{2}\). [web:3]

  • Compute approximate values: \(\sqrt5\approx 2.236\), so \(a\approx1.618>1\) and \(b\approx0.618\), hence \(0
  • Since \(a>1\), the term \(a^n\) grows without bound as n increases, while \(b^n\to0\) because \(0

Consider the expression \(L_n=\frac{a^n+b^n}{a^n-b^n}\). Divide the numerator and denominator by \(a^n\) (nonzero) to get \(L_n=\frac{1+(\frac{b}{a})^n}{1-(\frac{b}{a})^n}\). [web:3]

Because \(0<\frac{b}{a}<1\), the term \((\frac{b}{a})^n\to0\) as \(n\to\infty\), giving \(\lim_{n\to\infty}L_n=\frac{1+0}{1-0}=1\). [web:3]

Explanation of each option

Option (A): 1

Both numerator and denominator are dominated by \(a^n\) as n grows, so their ratio tends to \(\frac{a^n}{a^n}=1\), making option (A) correct. [web:3]

Option (B): 1/2

For the limit to be 1/2, the numerator would have to grow asymptotically half as fast as the denominator, but both behave like \(a^n\), so the ratio tends to 1, not 1/2. [web:3]

Option (C): 0

A limit of 0 would require the numerator to become negligible compared with the denominator, yet both contain the same dominant term \(a^n\), so the ratio cannot approach 0. [web:3]

Option (D): does not exist

Rewriting in terms of \((\frac{b}{a})^n\) shows that the expression converges smoothly to \(\frac{1+0}{1-0}\), so the limit exists and is finite, ruling out this option. [web:3]

 

Leave a Reply

Your email address will not be published. Required fields are marked *

Latest Courses