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

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

Don't understand how to approach this question

• Is there any reason for the limit to exist? – Lost Dec 12 '13 at 2:20
• @Lost What do you mean? – user115174 Dec 12 '13 at 2:21
• As $x\to -\infty$, what happens to the first term? The second term? What can you conclude? – user7530 Dec 12 '13 at 2:21
• I mean can you find any indication that this limit won't blow up? – Lost Dec 12 '13 at 2:22
• 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 2:46

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.

• Thanks for the edit... Poorly typed answer... – Umberto Dec 18 '13 at 7:21

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.

• 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 9:36
• 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
• Ah, okay, gotcha. – user1729 Dec 24 '13 at 13:58

Is it strictly enough?

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

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

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.

• Thanks for your help Suzanne but I'm supposed to be able to do this problem without a calculator – user115174 Dec 12 '13 at 2:47
• @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 3:00
• @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 3:05