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The values of constants $\mathrm{a}$ and $\mathrm{b}$, so that $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0$ are
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The correct answer is:
$a=1, b=-1$
$$
\begin{array}{l}
\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0 \\
\Rightarrow \lim _{x \rightarrow \infty}\left[\frac{x^{2}(1-a)-(a+b) x-b+1}{x+1}\right]=0 \\
\Rightarrow 1-a=0 \text { and } a+b=0 \Rightarrow a=1, b=-1
\end{array}
$$
\begin{array}{l}
\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0 \\
\Rightarrow \lim _{x \rightarrow \infty}\left[\frac{x^{2}(1-a)-(a+b) x-b+1}{x+1}\right]=0 \\
\Rightarrow 1-a=0 \text { and } a+b=0 \Rightarrow a=1, b=-1
\end{array}
$$
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