Question

If $\underset{x⟶\infty }{lim}(\frac{x^2+x+1}{x+1}-ax-b)=4$ then

• A. $a=1,b=4$
• B. $a=1,b=-4$
• C. $a=2,b=-3$
• D. $a=2,b=3$

Answer 1

Answer: (B)

$a=1,b=-4$

We have,

${lim}_{x\to \infty }(\frac{x^2+x+1}{x+1}-ax-b)=4$

${lim}_{x\to \infty }(x+\frac{1}{x+1}-ax-b)=4$

${lim}_{x\to \infty }(\frac{1}{x+1}+(1-a)x-b)=4$

As limit is finite, hence coefficient of $x$ should be zero.

$\Rightarrow 1-a=0$

$\Rightarrow a=1$

$\Rightarrow 4={lim}_{x\to \infty }\{-b\}$

$\Rightarrow 4=-b$

$\Rightarrow b=-4$

Hence, this is the answer.