Discussion Forum

if  $\lim_{x \rightarrow \infty}\left(\frac{x^{2}-x+1}{x+1}-ax-b\right)=4$ , then

Answer:

Explanation:

Concept involved   $( \frac{\infty}{\infty})$ form

 $\lim_{x \rightarrow \infty}\frac{a_{0}x^{n}+a_{1}x^{n-1}+.....+a_{n}}{b_{0}x^{m}+b_{1}x^{m-1}+.....+b_{m}}$

=$\begin{cases}0 ,& if n=m\\\frac{a_{0}}{b_{0}}, & if n <m \\+\infty, &if n>m and a_{0}b_{0}>0\\-\infty&if n>m and a_{0}b_{0}<0\end{cases}$

Description of Situation As to make degree of numerator equal to degree of denominator.

Sol. $\lim_{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-ax-b\right)=4$

$\Rightarrow$   $\lim_{x \rightarrow \infty}\frac{x^{2}+x+1-ax^{2}-ax-bx-b}{x+1}=4$

$\Rightarrow$  $\lim_{x \rightarrow \infty}\frac{x^{2}(1-a)+x(1-a-b)+(1-b)}{x+1}=4$

 Here, we make degree of N^r = degree of D^r

$\therefore$   1-a=0

 and  $\lim_{x \rightarrow \infty}\frac{x(1-a-b)+(1-b)}{x+1}=4$

$\Rightarrow$    1-a-b=4

$\Rightarrow$   b=-4   $[ \because (-a)=0]$