Question

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

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

Answer 1

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

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

$\underset{x\to \infty }{lim}\left(\frac{x^2(1-a)+x(1-a-b)+(1-b)}{x+1}\right)=4$

Limit is finite , It exists when $1-a=0\Rightarrow a=1$

then $\underset{x\to \infty }{lim}\left(\frac{1-a-b+\frac{1-b}{x}}{1+\frac{1}{x}}\right)=4$

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