Find the real parameters $a$ and $b$ in limit Can you please help me solve this problem which requires to find the values of real parameters $a$ and $b$ so that the relation below is satisfied:
$$\lim_{x\to -\infty}(\sqrt {x^2+x+1}-ax+b)=0$$
Thank you very much!

You have
$$
\lim_{x\to -\infty}\left(\sqrt {x^2+x+1}-ax+b\right)=0.
$$
Let $x=-y$, then you have
$$
\begin{align}
\lim_{-y\to -\infty}\left(\sqrt {(-y)^2+(-y)+1}-a(-y)+b\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}+ay+b\right)&=0.\tag1
\end{align}
$$
Now, divide both sides of $(1)$ by $y$. It turns out to be
$$
\begin{align}
\lim_{y\to \infty}\left(\frac{\sqrt {y^2-y+1}+ay+b}{y}\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {\frac{y^2-y+1}{y^2}}+a+\frac{b}{y}\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {1-\frac{1}{y}+\frac{1}{y^2}}+a+\frac{b}{y}\right)&=0\\
1+a&=0\\
a&=\boxed{\Large\color{blue}{-1}}
\end{align}
$$
Hence, $(1)$ becomes
$$
\begin{align}
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}-y+b\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}-y\right)=-b\\
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}-\sqrt{y^2}\right)=-b\\
\lim_{y\to \infty}\left(\left(\sqrt {y^2-y+1}-\sqrt{y^2}\right)\cdot\frac{\sqrt {y^2-y+1}+\sqrt{y^2}}{\sqrt {y^2-y+1}+\sqrt{y^2}}\right)=-b\\
\lim_{y\to \infty}\left(\frac{y^2-y+1-y^2}{\sqrt {y^2-y+1}+\sqrt{y^2}}\right)=-b\\
\lim_{y\to \infty}\left(\frac{-y+1}{\sqrt {y^2-y+1}+\sqrt{y^2}}\right)=-b.\tag2
\end{align}
$$
Now, divide the numerator and denominator part in $(2)$ by $y$, yield
$$
\begin{align}
\lim_{y\to \infty}\left(\frac{-1+\frac{1}{y}}{\sqrt {1-\frac{1}{y}+\frac{1}{y^2}}+\sqrt{1}}\right)&=-b\\
\frac{-1+0}{\sqrt {1-0+0}+\sqrt{1}}&=-b\\
-\frac{1}{2}&=-b\\
b&=\boxed{\Large\color{blue}{\frac{1}{2}}}
\end{align}
$$
