Limit of $\sqrt{x^2-6x+7}-x$ as x approaches negative infinity

Question

What is $\lim\limits_{x\to-\infty}(\sqrt{x^2-6x+7}-x)$ ?

Don't understand how to approach this question

As $x\to -\infty$, what happens to the first term? The second term? What can you conclude? — user7530, Dec 12 '13 at 02:21
I mean can you find any indication that this limit won't blow up? — Lost, Dec 12 '13 at 02:22
I tried rationalizing it but don't know if I'm doing the right steps — user115174, Dec 12 '13 at 02:38
Your first step should be to check whether or not the limit produces an indeterminant form — JessicaK, Dec 12 '13 at 02:41
Ok so I take the conjugate and I get -6x + 7 / (sqrt(x^2-6x+7) + x), I divide by the highest degree and I get 6 / -1 + 1, which gives me 6/0... does that mean the answer is infinity? — user115174, Dec 12 '13 at 02:46
@user115174 : with expressions like "6/0", you have to be careful. You don't always get infinity. Is the denominator positive? Negative? Changing sign? — Stefan Smith, Dec 12 '13 at 03:01
@StefanSmith But in this instance could the "6/0" be concluded to equal infinity? If it was something like "-6/0", would that equal negative infinity? — user115174, Dec 12 '13 at 03:06
@user115174 The problem would be much more interesting if the limit is x to infinity — imranfat, Dec 12 '13 at 03:09
You can find several questions of similar type on this site: http://math.stackexchange.com/questions/512768/explanation-for-lim-x-to-infty-sqrtx2-4x-x-2-and-not-0 http://math.stackexchange.com/questions/415853/the-limit-of-lim-limits-x-to-infty-sqrtx23x-4-x http://math.stackexchange.com/questions/580619/evaluating-lim-limits-x-to-infty-sqrtx2-3x5-sqrtx22x1 Have a look there. Can you use similar methods for your limit? — Martin Sleziak, Dec 12 '13 at 11:45
Since it seems from your comments that you already have partial solution, you should post it (perhaps with more details) as an answer or edit it into your question, so that people can comment on your solution. (Of course, you can only do this after the question is reopened.) — Martin Sleziak, Dec 12 '13 at 11:51
As the first two comments suggest, this could in fact be solved easily without rationalizing the fractions. — Martin Sleziak, Dec 12 '13 at 12:14
@MartinSleziak While I appreciate everyone's help,I feel like everyone on this site needs to relax if someone's math skills aren't up to par with theirs.I realize I should've been more thorough in searching past questions similar to mine, but the effort and time some of the people put in trying to "teach" me could've have easily been translated in them answering the questions I had. Discouragement isn't a proper form of teaching. I'm asking the question because I don't know it, answering by saying "this could in fact be solved "EASILY" without rationalizing the functions isn't too encouraging. — user115174, Dec 12 '13 at 17:06
The reason why I made that comment was that I've given you several links to questions, where rationalizing is used for computing some limit. And I have added the comment to explain that in fact it is not necessary to rationalize in this case to get the limit. As I have made the same mistake (immediately trying to rationalize to expression) I definitely do not have any right to comment on your math skills and if you interpreted my comment in this way, then it is definitely not what I intended. — Martin Sleziak, Dec 12 '13 at 17:15
probably an abstract duplicate of http://math.stackexchange.com/questions/30040 (with $n = 2$) — Ben, Dec 17 '13 at 07:06
@Ben I think that $x\to\infty$ and $x\to-\infty$ makes a difference here. (Notice that the answers in the link you have given are finite limits, the answer to this question is $+\infty$.) But I think that the link you have added might be useful for the OP, anyway. (I did not notice this at first; that's why I have also provided several links to similar limits, with $x\to\infty$, in my comment above.) — Martin Sleziak, Dec 17 '13 at 13:10
@Martin Oops—didn't even notice the minus! No harm done nonetheless. (I definitely agree with @$ $imranfat's comment) — Ben, Dec 17 '13 at 13:26
A similar limit posted recently: http://math.stackexchange.com/questions/627427/puzzled-by-displaystyle-lim-x-to-infty-sqrtx2x-x — Martin Sleziak, Jan 05 '14 at 09:54

score 4 · Answer 1 · edited Dec 17 '13 at 13:06

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Mainly the idea is the following: $x^2$ asymptotically (for $x\to \infty$) goes faster to $\infty$ than $x$. That means that in your square root you can neglect $-6x+7$. So your limit is $$ \lim\limits_{x\to-\infty}(\sqrt{x^2-6x+7}-x)=\lim\limits_{x\to-\infty}(\sqrt{x^2}-x)=\lim\limits_{x\to-\infty}(|x|-x)=\lim\limits_{x\to+\infty}2|x|=\infty $$ I hope is clearer now.

edited Dec 17 '13 at 13:06

Martin Sleziak

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answered Dec 17 '13 at 07:31

Umberto

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Thanks for the edit... Poorly typed answer... – Umberto Dec 18 '13 at 07:21

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

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Notice that $\lim\limits_{x\to-\infty} x^2-6x+7=\infty$. (If $P(x)=x^n+\dots+a_1x+a_0$ is a monic polynomial of even degree then $\lim\limits_{x\to-\infty} P(x)=\lim\limits_{x\to\infty}P(x)=\infty$.)
Therefore also $\lim\limits_{x\to-\infty} \sqrt{x^2-6x+7}=\infty.$

Thus we get
$$\lim\limits_{x\to-\infty} (\sqrt{x^2-6x+7}-x)=\lim\limits_{x\to-\infty} \sqrt{x^2-6x+7}+ \lim\limits_{x\to-\infty} (-x)=\infty+\infty=\infty.$$

You have mentioned in your comment that you have tried rationalization. If you do it this way, you get $$\sqrt{x^2-6x+7}-x = \frac{-6x+7}{\sqrt{x^2-6x+7}+x} = \frac{-6+\frac7x}{-\sqrt{1-\frac6x+\frac7{x^2}}+1}$$ since for negative $x$ we have $-a=|a|=\sqrt{a^2}$.

This means that you get limit of the type $\frac60$. In general, we can't say anything about expressions like this. (For example $\lim\limits_{t\to0} \frac 6{t^2}=\infty$ but $\lim\limits_{t\to0} \frac6t$ does not exist.)

But in this case you have the additional information that the denominator is non-negative. So it would be more precise to say that this is a limit of the type $\frac6{0^+}$, which is indeed $+\infty$. (If the numerator has positive limit and the denominator is always positive and tends to zero, then the limit of the fraction is $+\infty$.)

Limits of this type are also briefly mentioned in the Wikipedia article on indeterminate forms. (I will also add a link to recent revision, which should work in the future even if the article is substantially changed.

edited Apr 13 '17 at 12:21

Community

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answered Dec 17 '13 at 15:19

Martin Sleziak

53,687

Aren't you assuming that the limits exist when you go $\lim\limits_{x\to-\infty} (\sqrt{x^2-6x+7}-x)=\lim\limits_{x\to-\infty} \sqrt{x^2-6x+7}+ \lim\limits_{x\to-\infty} (-x)$? (As Cameron Williams said to Porfiriy's answer.) – user1729 Dec 24 '13 at 09:36
Whenever I write something like $\lim\limits_{x\to a} (f(x)+g(x))=\lim\limits_{x\to a}f(x)+\lim\limits_{x\to a}g(x)$, this is intended as a shortcut for If the limits on the RHS of this equality exist, then so does the limit on the LHS and the value of this limit is equal to their sum. This result is also true for limits of the type $\infty=\infty+\infty$. (We use the addition on extended real line.) – Martin Sleziak Dec 24 '13 at 12:03
Yes, but if $\lim f+\lim g$ does not exist then we cannot conclude that $\lim (f+g)$ does not exist. But this is what you are doing, no? – user1729 Dec 24 '13 at 13:28
No, what I claim is that if $\lim f=+\infty$ and $\lim g=+\infty$, then also $\lim (f+g)=+\infty$. – Martin Sleziak Dec 24 '13 at 13:56
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Ah, okay, gotcha. – user1729 Dec 24 '13 at 13:58

score 1 · Answer 3 · answered Dec 17 '13 at 07:46

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Is it strictly enough?

$$\lim\limits_{x\to-\infty}(\sqrt{x^2-6x+7}-x)=\lim\limits_{x\to-\infty}(\sqrt{x^2-6x+7}) + \lim\limits_{x\to-\infty}(-x)=\lim\limits_{x\to-\infty}(|x|\sqrt{1-\frac{6}{x}+\frac{7}{x^2}}) + \infty=\lim\limits_{x\to-\infty}(|x|\sqrt{1-0+0}) + \infty=\lim\limits_{x\to-\infty}(|x|) + \infty=\infty + \infty = \infty $$

answered Dec 17 '13 at 07:46

Porfiriy

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You cannot break apart limits like that if they don't exist. – Cameron Williams Dec 17 '13 at 15:45

score 0 · Answer 4 · answered Dec 12 '13 at 02:46

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Taking Calculus many years ago, I recall taking an equation similar to this one and if finding myself confused, plugging this expression into a calculator and see how it behaves. When I plugged it into my calculator I then looked towards negative infinity since that is what the limit is requesting and see the function increasing. You can confirm this by plugging in values of x and observing the output. I started with -10, then -100, -10000, and finally -1000000000. Try this method and see what you can infer from the results.

answered Dec 12 '13 at 02:46

Suzanne Frank

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Thanks for your help Suzanne but I'm supposed to be able to do this problem without a calculator – user115174 Dec 12 '13 at 02:47
@user115174 : use a calculator in the privacy of your own dorm/apartment/whatever without telling anyone or writing anything down. You will see a pattern and that tells you what the answer is. Then prove it is the right answer without relying on a calculator. No one will know you used a calculator! – Stefan Smith Dec 12 '13 at 03:00
@StefanSmith Ok, I understand that the function goes to infinity as this limit approaches negative infinity, but it is the proving part which I am asking about. If I am asked this on a quiz, I would have to be able to prove it mathematically and show steps of my work, which I have no idea how to do.... – user115174 Dec 12 '13 at 03:05

Limit of $\sqrt{x^2-6x+7}-x$ as x approaches negative infinity

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