How do I evaluate $\displaystyle \lim_{x \to -\infty}\left(\sqrt{x^2-8x+1}-x\right)$? As asked in the title.
I want to know how can I show that the value approaches to positive infinity as x approaches to negative infinity,without looking at its graph.
Besides, I know that I can use the formula $\displaystyle (a+b)(a-b)=a^2-b^2$, but then I don't know what should I do next.
 A: $\lim\limits_{x\to-\infty}(\sqrt{x^2-8x+1}-x)$
$=\lim\limits_{x\to-\infty}(\sqrt{x^2-8x+1}-x)\left(\dfrac{\sqrt{x^2-8x+1}+x}{\sqrt{x^2-8x+1}+x}\right)$
$=\lim\limits_{x\to-\infty}\dfrac{-8x+1}{\sqrt{x^2-8x+1}+x}$
Divide top and bottom by $x$. When dividing the bottom by $x$, use $x=-\sqrt{x^2}$. (There is a negative sign, because $x\to -\infty$ so $x<0$.)
$=\lim\limits_{x\to-\infty}\dfrac{-8+\dfrac{1}{x}}{-\sqrt{1-\dfrac{8}{x}+\dfrac{1}{x^2}}+1}$
$=\lim\limits_{x\to-\infty}\dfrac{-8+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x}\left(8-\dfrac{1}{x}\right)}+1}$
For large negative $x$, the part inside the square root is a little more than $1$. So the denominator is small and negative. Since the numerator is close to $-8$, the whole thing is large positive.
So the limit is $+\infty$, or you could say the limit does not exist.
A: Another approach is to change variables: $x \mapsto -x$. Your problem then becomes
$$\lim_{x \rightarrow \infty} \big( \sqrt{x^2 + 8x +1} + x\big).$$
If we consider only $x > 0$ (which is fine as you are taking the limit as $x \rightarrow \infty$) you get
$$\lim_{x \rightarrow \infty}\big(\sqrt{x^2 + 8x +1} + x\big) > \lim_{x \rightarrow \infty}\sqrt{x^2} = \lim_{x \rightarrow \infty}x.$$
The inequality is justified as the left hand side is an increasing function. By comparison, since the right hand side tends to infinity, so must the left hand side.
