<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] as far as understanding, you have, in essence,
$$ \left| x - \frac{5}{2} \right|   - x  .  $$
for large positive $x$ you get close to $-5/2,$ but for large negative $x,$ meaning $x$ is negative and $|x|$ is large, you have roughly 
$$ 2 |x| + \frac{5}{2}  $$
which grows without bound. For example, if $x = -10,$ the original expression becomes
$$ \sqrt{100 - (-50) + 1} - (-10) = \sqrt {151} + 10 \approx 22.2882 $$
If $x = -100,$
$$ \sqrt{10000 - (-500) + 1} - (-100) = \sqrt {10501} + 100 \approx 202.474 $$
If $x = -1000,$
$$ \sqrt{1000000 - (-5000) + 1} - (-1000) = \sqrt {1005001} + 1000 \approx 2002.49738 $$
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