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If lim⁡x→∞[x2+1x+1−ax−b]=0\lim_{x \to \infty} \left[ \frac{x^2 + 1}{x + 1} - ax - b \right] = 0x→∞lim​[x+1x2+1​−ax−b]=0, then (a,b) is ​
Question

If limx[x2+1x+1axb]=0\lim_{x \to \infty} \left[ \frac{x^2 + 1}{x + 1} - ax - b \right] = 0, then (a,b) is ​

A.

(1 ,1)

B.

(2, –1)

C.

(1, –1)

D.

(1, 0)

Correct option is C

Given: limx[x+1x2+1axb]=0Step 1: Polynomial divisionx+1x2+1=x1+2x+1So, limx[x1+2x+1axb]=0Group terms: =(1a)x+(1b)+2x+1For the limit to be 0 as x:1a=0=>a=11b=0=>b=1Final Result: (a,b)=(1,1)\text{Given: } \lim_{x \to \infty} \left[ \frac{x+1}{x^2 + 1} - ax - b \right] = 0 \\[8pt]\text{Step 1: Polynomial division} \\\frac{x+1}{x^2 + 1} = x - 1 + \frac{2}{x+1} \\[6pt]\text{So, } \lim_{x \to \infty} \left[ x - 1 + \frac{2}{x+1} - ax - b \right] = 0 \\[6pt]\text{Group terms: } \\= (1 - a)x + (-1 - b) + \frac{2}{x+1} \\[6pt]\text{For the limit to be 0 as } x \to \infty: \\1 - a = 0 \Rightarrow a = 1 \\-1 - b = 0 \Rightarrow b = -1 \\[6pt]\text{Final Result: } \boxed{(a, b) = (1, -1)}​​

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