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
 A: 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.
A: 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.
A: Is it strictly enough?
$$\lim\limits_{x\to-\infty}(\sqrt{x^2-6x+7}-x)=\lim\limits_{x\to-\infty}(\sqrt{x^2-6x+7}) + \lim\limits_{x\to-\infty}(-x)=\lim\limits_{x\to-\infty}(|x|\sqrt{1-\frac{6}{x}+\frac{7}{x^2}}) + \infty=\lim\limits_{x\to-\infty}(|x|\sqrt{1-0+0}) + \infty=\lim\limits_{x\to-\infty}(|x|) + \infty=\infty + \infty = \infty
$$
A: 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.
