What is \(\rm \displaystyle\lim_{n \rightarrow \infty} \frac{a^n+b^n}{a^n-b^n}\) where a > b > 1, equal to?

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  1. -1
  2. 0
  3. 1
  4. Limit does not exist

Answer (Detailed Solution Below)

Option 3 : 1
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Detailed Solution

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Given:

f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n+b^n}{a^n-b^n}\) and

a > b > 1

Calculation:

We have,

a > b > 1

⇒ \(\frac{a}{b}>1\) or \(\frac{b}{a}<1\)

Given that,

f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n+b^n}{a^n-b^n}\)    

f(x) \(\rm \displaystyle\lim_{n → ∞} \frac{a^n[1+(\frac{b}{a}) ^{n}]}{a^n[1 - (\frac{b}{a}) ^{n}]}\) 

⇒ f(x) \(\rm \displaystyle\lim_{n → ∞} \frac{[1+(\frac{b}{a}) ^{n}]}{[1 - (\frac{b}{a}) ^{n}]}\) 

Taking limit n→∞

⇒ f(x) \(\rm \displaystyle\lim_{n → ∞} \frac{[1+(\frac{b^{\infty}}{a^{\infty}}) ]}{[1 - (\frac{b^{\infty}}{a^{\infty}}) ]}\)  

⇒ f(x) = \(\rm \frac{1 + 0}{1 - 0}\)       (∵ \(\frac{b}{a}<1\) )

∴  f(x) = 1

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