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BITSAT 2017
Exam
BITSAT
Year
2017
Subject
Mathematics
Chapter
Limits

BITSAT 2017 · Mathematics · Limits

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

If lim⁡n→∞(an−1+n21+n)=b\lim_{n \to \infty}\left(an - \dfrac{1 + n^{2}}{1 + n}\right) = b, where aa is finite number, then

Options

  1. Aa=2a = 2
  2. Ba=0a = 0
  3. Cb=1b = 1
  4. Db=−1b = - 1

Correct answer

C. b=1b = 1

Step-by-step solution

lim⁡n→∞an(1+n)−(1+n2)1+n\lim_{n \to \infty}\dfrac{an ( 1 + n ) - \left(1 + n^{2}\right)}{1 + n}

=lim⁡n→∞(a−1)n2+an−1n+1=∞= \lim_{n \to \infty}\dfrac{( a - 1 ) n^{2} + an - 1}{n + 1} = \infty,

If a−1≠0a - 1 \ne 0 limit does not exist and if a−1=0a - 1 = 0, then

lim⁡n→∞an−1n+1=a=b\lim_{n \to \infty}\dfrac{an - 1}{n + 1} = a = b

⇒a=b=1⇒ a = b = 1

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