If $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4$, then

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Limits

lim_{x \to \infty} (\frac{x ^{2} + x + 1}{x + 1} - a x - b) = 4

, then

Answer: (B)

lim_{x \to \infty} \frac{x ^{2} + x + 1 - a x ^{2} - a x - b x - b}{x + 1} = 4

. This gives

lim_{x \to \infty} \frac{( 1 - a ) x ^{2} + ( 1 - a - b ) x + ( 1 - b )}{x + 1} = 4

. For the limit to be finite, we need

1 - a = 0

, so

a = 1

. Then

lim_{x \to \infty} \frac{( 1 - 1 - b ) x + ( 1 - b )}{x + 1} = lim_{x \to \infty} \frac{- b x + ( 1 - b )}{x + 1} = - b = 4

, so

b = - 4

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option