Question

If $\underset{x\to \mathrm{\infty }}{lim}(\frac{x^2+x+1}{x+1}-px-q)=-3$, then $p$ and $q$ is:

• (A) p = 1, q = 3
• (B) p = 2, q = 3
• (C) p = 3, q = 1
• (D) p = 3, q = 2

Answer 1

The Correct Option is A

Explanation:
Given: $\underset{x\to \mathrm{\infty }}{lim}(\frac{x^2+x+1}{x+1}-px-q)=-3$On simplifying, we get:$\Rightarrow \underset{x\to \mathrm{\infty }}{lim}\left(\frac{x^2+x+1-p{x}^2-px-qx-q}{x+1}\right)=-3$$\Rightarrow \underset{x\to \mathrm{\infty }}{lim}\left(\frac{x^2(1-p)+x(1-p-q)+1-q}{x+1}\right)=-3$  As we can see limit gives finite value. So, this is possible only when the coefficient of higher degree term will be zero.Therefore, $(1-p)=0$Or, $p=1$$\Rightarrow \underset{x\to \mathrm{\infty }}{lim}\left(\frac{x(1-p-q)+1-q}{x+1}\right)=-3$ ....(1)Dividing and multiplying by $x$ in both numerator and denominator, we get:$\Rightarrow \underset{x\to \mathrm{\infty }}{lim}\left(\frac{x[(1-p-q)+\frac{(1-q)}{x}]}{x[1+\frac{1}{x}]}\right)=-3$$\Rightarrow \underset{x\to \mathrm{\infty }}{lim}\left(\frac{[(1-p-q)+\frac{(1-q)}{x}]}{[1+\frac{1}{x}]}\right)=-3$$\Rightarrow \left(\frac{[(1-p-q)+0]}{[1+0]}\right)=-3$$\Rightarrow (1-p-q)=-3$ $\Rightarrow (1-1-q)=-3(\because p=1)$$\therefore q=3$Hence, the correct option is (A).