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

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

Vedclass · Accepted Answer

$a=1, b=-4$

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(B) Given $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-ax-b\right)=4$. Simplify the expression inside the limit: $\frac{x^2+x+1 - (ax+b)(x+1)}{x+1} = \frac{x^2+x+1 - (ax^2+ax+bx+b)}{x+1} = \frac{x^2(1-a) + x(1-a-b) + (1-b)}{x+1}$. For the limit to be a finite value as $x \rightarrow \infty$,the coefficient of the highest power of $x$ in the numerator must be zero. Thus,$1-a=0$,which gives $a=1$. Substituting $a=1$ into the expression: $\lim _{x \rightarrow \infty} \frac{x(1-1-b) + (1-b)}{x+1} = \lim _{x \rightarrow \infty} \frac{-bx + (1-b)}{x+1}$. Dividing numerator and denominator by $x$: $\lim _{x \rightarrow \infty} \frac{-b + \frac{1-b}{x}}{1 + \frac{1}{x}} = -b$. Given the limit is $4$,we have $-b=4$,so $b=-4$. Therefore,$a=1$ and $b=-4$.

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

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