If limn→∞an-(1+n^2)/(1+n)=b, where a is finite number, then

Correct answer: (C) b=1. limn→∞(an(1+n)-1+n^2)/(1+n) =limn→∞((a-1)n^2+an-1)/(n+1)=∞, If a-1≠0 limit does not exist and if a-1=0, then limn→∞(an-1)/(n+1)=a=b ⇒ a=b=1

Q72 / 125 · BITSAT 2017 · Mathematics

If $\lim_{n \to \infty} (a n - \frac{1 + n^{2}}{1 + n}) = b$ , where $a$ is finite number, then

A
$a = 2$
B
$a = 0$
Correct answer
C
$b = 1$
D
$b = - 1$

Solution

$\lim_{n \to \infty} \frac{an (1 + n) - (1 + n^{2})}{1 + n}$

$= \lim_{n \to \infty} \frac{(a - 1) n^{2} + a n - 1}{n + 1} = \infty$ ,

If $a - 1 \neq 0$ limit does not exist and if $a - 1 = 0$ , then

$\lim_{n \to \infty} \frac{a n - 1}{n + 1} = a = b$

$\Rightarrow a = b = 1$

Hence, the correct answer is option: 'C'.

If limn→∞an-1+n21+n=b, where a is finite number, then

Solution

More Limits Questions

If $\lim_{n \to \infty} (a n - \frac{1 + n^{2}}{1 + n}) = b$ , where $a$ is finite number, then