Question

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

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

Answer 1

The correct option is B $a=1,b=-4$

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

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

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

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

As the limit is infinite hence its coefficient of $x=0$

$\Rightarrow 1-a=0$ or $a=1$

$\Rightarrow {lim}_{x\to \infty }(\frac{1}{x(1+\frac{1}{x})}-b)=4$ since ${lim}_{x\to \infty }\frac{1}{x}=0$

$\Rightarrow -b=4$ or $b=-4$

$\therefore a=1,b=-4$