\(\rm \displaystyle\lim_{n \rightarrow \infty} \frac{a^n+b^n}{a^n-b^n}\) किसके बराबर है, जहाँ a > b > 1 है?

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NDA 02/2021: Maths Previous Year paper (Held On 14 Nov 2021)

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Answer 1

दिया गया है:

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

a > b > 1

गणना:

हमारे पास है,

a > b > 1

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

दिया गया है कि,

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

सीमा n→∞ लेने पर

⇒ 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