Correct Answer - Option 3 : 1
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