If $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$, then the ordered pair $(a, b)$ is :

(C) $\left(1,-\frac{1}{2}\right)$

(A) $\left(-1, \frac{1}{2}\right)$

(B) $\left(-1,-\frac{1}{2}\right)$

(D) $\left(1, \frac{1}{2}\right)$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Limits

lim_{x \to \infty} (x^{2} - x + 1 - a x) = b

, then the ordered pair

(a, b)

is :

Answer: (C)

For the limit to exist as

x \to \infty

, we need

a = 1

(otherwise the expression diverges). With

a = 1

, rationalize:

x^{2} - x + 1 - x = \frac{( x ^{2} - x + 1 ) - x ^{2}}{x ^{2} - x + 1 + x} = \frac{- x + 1}{x ^{2} - x + 1 + x}

. Dividing numerator and denominator by

x

\frac{- 1 + 1/ x}{1 - 1/ x + 1/ x ^{2} + 1} \to \frac{- 1}{1 + 1} = - \frac{1}{2}

x \to \infty

. Thus

(a, b) = (1, - 1/2)

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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