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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!

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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 $$ \begin{align} \lim_{y\to \infty}\left(\frac{-1+\frac{1}{y}}{\sqrt {1-\frac{1}{y}+\frac{1}{y^2}}+\sqrt{1}}\right)&=-b\\ \frac{-1+0}{\sqrt {1-0+0}+\sqrt{1}}&=-b\\ -\frac{1}{2}&=-b\\ b&=\boxed{\Large\color{blue}{\frac{1}{2}}} \end{align} $$

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let $-x=t$,then $$\Longrightarrow \lim_{t\to +\infty}(\sqrt{t^2-t+1}+at+b)=0$$ $$\Longrightarrow\lim_{t\to\infty}\dfrac{\sqrt{t^2-t+1}+at+b}{t}=0$$ $$\Longrightarrow\lim_{t\to\infty}\left(\dfrac{\sqrt{t^2-t+1}}{t}+a+\dfrac{b}{t}\right)=0$$

$$\Longrightarrow a=-\lim_{t\to \infty}\left(\dfrac{\sqrt{t^2-t+1}}{t}+\dfrac{b}{t}\right)=-1+0=-1$$ so $$b=-\lim_{t\to\infty}(\sqrt{t^2-t+1}-t)=\dfrac{1}{2}$$

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  • $\begingroup$ Thank you,but there is a problem, I cannot understand your explanation,though I see it's effective. Can you please give me a more exact one? $\endgroup$ – Ivan Gandacov Apr 11 '14 at 16:51
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    $\begingroup$ Still there is one more question. Is it possible for you to comment why have you put in the second line $t$ at the denominator? $\endgroup$ – Ivan Gandacov Apr 11 '14 at 17:08
  • $\begingroup$ use this:if $\lim_{x\to +\infty}f(x)=0$,then we have $\lim_{x\to\infty}\dfrac{f(x)}{x}=0$ $\endgroup$ – math110 Apr 11 '14 at 17:13
  • $\begingroup$ Thank you very much, but this question remains for me unclear. I still do not understand why have you put that "t". $\endgroup$ – Ivan Gandacov Apr 11 '14 at 17:25
  • $\begingroup$ Sorry for my being so dumb, but is this a property of limits? $\endgroup$ – Ivan Gandacov Apr 11 '14 at 17:31
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Hint: Note that you want $$(ax-b)^2=a^2x^2-2ab\cdot x+b^2$$ to closely resemble $$\left(\sqrt{x^2+x+1}\right)^2=x^2+x+1.$$

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  • $\begingroup$ Thanks. Can you suggest the next step? $\endgroup$ – Ivan Gandacov Apr 11 '14 at 17:19

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