TITLE: How do I evaluate $\displaystyle \lim_{x \to -\infty}\left(\sqrt{x^2-8x+1}-x\right)$?
QUESTION [3 upvotes]: 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.

REPLY [2 votes]: 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.
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