Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 does one solve this?

share|cite|improve this question

4 Answers 4

up vote 5 down vote accepted

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 $$\lim_{x\to-\infty}\frac{x+2}{\sqrt{x^2-x+2}}=\lim_{x\to-\infty}\frac{x}{-x}=-1$$

share|cite|improve this answer
suppose you are using lhospital to solve this. do you care that in denminator its ininfity minus infnity –  Bak1139 Jun 30 '14 at 12:44
The L'Hôpital's rule isn't suitable in this case! You have this alternative: $$\sqrt{x^2-x+2}=|x|\sqrt{1-\frac1x+\frac2{x^2}}$$ –  user63181 Jun 30 '14 at 12:50
i dont understand what youve done here –  Bak1139 Jun 30 '14 at 13:39
@Bak113 He just factored $\,x^2\,$ inside the radical: $$\sqrt{x^2-x+2}=\sqrt{x^2\left(1-\dfrac1x+\dfrac2{x^2}\right)}=|x|\sqrt{1- \dfrac {1} {x}+\dfrac{2}{x^2}}.$$ –  Hakim Jun 30 '14 at 16:20
that was a tricky one, thanks –  Bak1139 Jul 13 '14 at 15:11

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 $1$.

share|cite|improve this answer
thins is the algebra IS needed... –  Bak1139 Jun 30 '14 at 13:40
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 $|\cdot|$. –  Alex Jun 30 '14 at 13:45

$$=\frac{1+\frac{2}{x}}{\sqrt{1-\frac{1}{x}+\frac{2}{x^2}}}\rightarrow 1$$

(this is an answer to original question)

share|cite|improve this answer
be careful with $x \to \infty$ –  Alex Jun 30 '14 at 12:39
@Alex : i do not get what actually do you want him/her to be careful with.. –  Praphulla Koushik Jun 30 '14 at 12:40
My point is that for $x \to \infty$ and $x \to - \infty$ the solutions are different. –  Alex Jun 30 '14 at 12:42
you are doubling the entire fraction by $(1/x / 1/x)$, right? –  Bak1139 Jun 30 '14 at 12:50
@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. –  Rene Schipperus Jun 30 '14 at 13:01

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 $\sqrt{x^{2}}=|x|$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.