displaystyle lim n arrow ∞ (an+bn/an-bn), where 1

Question

displaystyle lim n arrow ∞ (an+bn/an-bn), where 1<b<a, is equal to (A) -1 (B) 1 (C) 0 (D) None of these

Tardigrade Admin · Accepted Answer

We have, 1 < b < a. ∴ 0<(b/a)<1 ⇒ displaystyle lim n arrow ∞((b/a))n=0 So, displaystyle lim n arrow ∞ (an+bn/an-bn) = displaystyle lim n arrow ∞ (1+((b/a))n/1-((b)a)n)=(1+0/1-0)=1

Q. $n \to \infty lim \frac{a ^{n} + b ^{n}}{a ^{n} - b ^{n}}$ , where $1 < b < a$ , is equal to

$- 1$

1

0

None of these

We have, $1 < b < a$ .
$∴ 0 < \frac{b}{a} < 1$
$\Rightarrow n \to \infty lim (\frac{b}{a})^{n} = 0$
So, $n \to \infty lim \frac{a ^{n} + b ^{n}}{a ^{n} - b ^{n}}$
$= n \to \infty lim \frac{1 + ( \frac{b}{a} ) ^{n}}{1 - ( \frac{b}{a} ) ^{n}} = \frac{1 + 0}{1 - 0} = 1$

Q. n→∞lim​an−bnan+bn​, where 1<b<a, is equal to

−1

1

0

None of these

We have, 1<b<a. ∴0<ab​<1 ⇒n→∞lim​(ab​)n=0 So, n→∞lim​an−bnan+bn​ =n→∞lim​1−(ab​)n1+(ab​)n​=1−01+0​=1

Q. $n \to \infty lim \frac{a ^{n} + b ^{n}}{a ^{n} - b ^{n}}$ , where $1 < b < a$ , is equal to

$- 1$

We have, $1 < b < a$ .
$∴ 0 < \frac{b}{a} < 1$
$\Rightarrow n \to \infty lim (\frac{b}{a})^{n} = 0$
So, $n \to \infty lim \frac{a ^{n} + b ^{n}}{a ^{n} - b ^{n}}$
$= n \to \infty lim \frac{1 + ( \frac{b}{a} ) ^{n}}{1 - ( \frac{b}{a} ) ^{n}} = \frac{1 + 0}{1 - 0} = 1$