Find constants $a$ and $b$ such that $\lim_\limits{x\to\infty}\left(\frac{x^2+1}{x+1}-ax-b\right)=0$.
My attempt: $\frac{x^2 + 1}{ x+1} -ax -b = \frac {x^2 +1 - (x +1)(ax+b)}{x+1}$ but then what?
2026-03-26 14:34:23.1774535663
Find constants $a$ and $b$ such that $\lim_\limits{x\to\infty}\left(\frac{x^2+1}{x+1}-ax-b\right)=0$.
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$\lim_\limits{x\to\infty}\left(\frac{x^2-1+2}{x+1}-ax+b\right)=\lim_\limits{x\to\infty}\left(\frac{x^2-1}{x+1}+\frac{2}{x+1}-ax-b\right)=\lim_\limits{x\to\infty}\left(\frac{(x+1)(x-1)}{x+1}+\frac{2}{x+1}-ax-b\right)=\lim_\limits{x\to\infty}\left(x-1+\frac{2}{x+1}-ax-b\right)=\lim_\limits{x\to\infty}\left(x-1-ax-b\right)$
From here we can find that $a=1$ and $b=-1$.