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In taking each of the limits
$$\lim_{x\to -\infty}\frac{x+2}{\sqrt {x^2-x+2}}\quad \text{ and } \quad \lim_{x\to \infty}\frac{x+2}{\sqrt {x^2-x+2}},$$
I find that both give the value $1$, although it should in fact be getting $-1$ and $1$, respectively. This however doesn't show from the calculations...how does one solve this?

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You're right! First notice that $$x^2-x+2\sim_\infty x^2$$ hence $$\frac{x+2}{\sqrt{x^2-x+2}}\sim_\infty\frac{x}{\sqrt{x^2}}=\frac{x}{|x|}$$ the result for $x\to+\infty$ is clear (and equals $1$) and for $-\infty$ we get $$\lim_{x\to-\infty}\frac{x+2}{\sqrt{x^2-x+2}}=\lim_{x\to-\infty}\frac{x}{-x}=-1$$

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  • $\begingroup$ suppose you are using lhospital to solve this. do you care that in denminator its ininfity minus infnity $\endgroup$ – Bak1139 Jun 30 '14 at 12:44
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    $\begingroup$ The L'Hôpital's rule isn't suitable in this case! You have this alternative: $$\sqrt{x^2-x+2}=|x|\sqrt{1-\frac1x+\frac2{x^2}}$$ $\endgroup$ – user63181 Jun 30 '14 at 12:50
  • $\begingroup$ i dont understand what youve done here $\endgroup$ – Bak1139 Jun 30 '14 at 13:39
  • $\begingroup$ @Bak113 He just factored $\,x^2\,$ inside the radical: $$\sqrt{x^2-x+2}=\sqrt{x^2\left(1-\dfrac1x+\dfrac2{x^2}\right)}=|x|\sqrt{1- \dfrac {1} {x}+\dfrac{2}{x^2}}.$$ $\endgroup$ – Hakim Jun 30 '14 at 16:20
  • $\begingroup$ that was a tricky one, thanks $\endgroup$ – Bak1139 Jul 13 '14 at 15:11
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It all boils down (simple algebra omitted, see other answers ) to showing what happens to $$ \lim_{x \to -\infty}\frac{x}{|x|} $$ Remember the definition of the absolute value: for $x< 0 \ f(x) = -x$. Therefore your limit becomes $\lim_{x \to - \infty} \frac{x}{-x} = -1$. For $x \to \infty$ it is of course $1$.

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  • $\begingroup$ thins is the algebra IS needed... $\endgroup$ – Bak1139 Jun 30 '14 at 13:40
  • $\begingroup$ no doubt it is; others gave you good ideas on how to handle the algebra of the problem (bound the denominator); I outlined the main idea on how to treat the $|\cdot|$. $\endgroup$ – Alex Jun 30 '14 at 13:45
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$$=\frac{1+\frac{2}{x}}{\sqrt{1-\frac{1}{x}+\frac{2}{x^2}}}\rightarrow 1$$

(this is an answer to original question)

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  • $\begingroup$ be careful with $x \to \infty$ $\endgroup$ – Alex Jun 30 '14 at 12:39
  • $\begingroup$ @Alex : i do not get what actually do you want him/her to be careful with.. $\endgroup$ – user87543 Jun 30 '14 at 12:40
  • $\begingroup$ My point is that for $x \to \infty$ and $x \to - \infty$ the solutions are different. $\endgroup$ – Alex Jun 30 '14 at 12:42
  • $\begingroup$ you are doubling the entire fraction by $(1/x / 1/x)$, right? $\endgroup$ – Bak1139 Jun 30 '14 at 12:50
  • $\begingroup$ @Bak1139 Yes that is what is going on. Multiply top and bottom by $\frac{1}{x}$ but in the case $-\infty$ when you square $x$ to bring it inside the root you loose the sign so you have to compensate for that. $\endgroup$ – Rene Schipperus Jun 30 '14 at 13:01
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Your fundamental problem arises regarding the issue of signs when dealing with $x\to -\infty$ and it can be handled most easily (without applying too much thought and in almost mechanical fashion) by putting $x=-t$ and then letting $t \to\infty$. Thus we have $$\lim_{x\to -\infty}\frac{x+2}{\sqrt{x^{2}-x+2}}=\lim_{t \to \infty}\frac{-t+2}{\sqrt{t^{2} + t +2}}=-1$$ Note that this approach totally avoids the hassle of dealing with $|x|$ and the understanding that $\sqrt{x^{2}}=|x|$.

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