<s>[INST] Finding $\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x$ results in loss of information Let $f(x) = \sqrt{x^2-5x+1}-x$
Find $\lim_{x\to\infty}f(x)$
$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$
$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{\dfrac{-5x+1}{x}}{\dfrac{\sqrt{x^2-5x+1}+x}{x}}$$
$$\lim_{x\to\infty} \dfrac{-5+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{5}{x}+\dfrac{1}{x^2}}+1}$$
From here, I know that $\lim_{x\to\infty} \dfrac{1}{x} = 0$, $\lim_{x\to\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to\infty} \dfrac{1}{x^2} = 0$
$$\lim_{x\to\infty} \dfrac{-5}{\sqrt{1}+1}$$
$$\lim_{x\to\infty} \dfrac{-5}{2}$$
$$\lim_{x\to\infty} f(x) = \dfrac{-5}{2}$$
Everything up to here seems fine. The issue is when I try to find $\lim_{x\to-\infty} f(x)$
I also know that $\lim_{x\to-\infty} \dfrac{1}{x} = 0$, $\lim_{x\to-\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to-\infty} \dfrac{1}{x^2} = 0$
This would make me conclude that $\lim_{x\to-\infty}f(x) = \dfrac{-5}{2}$.
However, this is not the case because $\lim_{x\to-\infty}f(x) = \infty$
Desmos view of $f(x)$
Why am I arriving to the wrong answer and how can I algebraically prove that the answer is $\infty$?
 [/INST] Note that for $x\to -\infty$


*

*$\sqrt{x^2-5x+1} \to +\infty$

*$-x \to +\infty$
and therefore
$$\sqrt{x^2-5x+1}-x \to +\infty$$
As an alternative by $x=-y\to -\infty$ with $y\to +\infty$ we have
$$\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x=\lim_{y\to +\infty}\sqrt{y^2+5y+1}+y \to +\infty$$
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