If lim _{x rightarrow infty}left(frac{x^2+x+1 | vedclass

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Question

(B) Given $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$.Simplify the expression inside the limit:$\lim _{x \rightarrow \infty

Vedclass · Accepted Answer

$a=1, b=-4$

Vedclass · Answer

(B) Given $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$.Simplify the expression inside the limit:$\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-ax(x+1)-b(x+1)}{x+1}\right)=4$$\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1-ax^2-ax-bx-b}{x+1}\right)=4$$\lim _{x \rightarrow \infty}\left(\frac{(1-a)x^2+(1-a-b)x+(1-b)}{x+1}\right)=4$For the limit to be finite,the coefficient of $x^2$ must be zero:$1-a=0 \Rightarrow a=1$.Now,substitute $a=1$ into the expression:$\lim _{x \rightarrow \infty}\left(\frac{(1-1-b)x+(1-b)}{x+1}\right)=4$$\lim _{x \rightarrow \infty}\left(\frac{-bx+(1-b)}{x+1}\right)=4$Divide numerator and denominator by $x$:$\lim _{x \rightarrow \infty}\left(\frac{-b+\frac{1-b}{x}}{1+\frac{1}{x}}\right)=4$$-b=4 \Rightarrow b=-4$.Thus,$a=1$ and $b=-4$.

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$,then:

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