<s>[INST] Why does trying to compute $\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}}$ result in the negative of the answer given? My textbook asks me to evaluate the limit $$\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}}$$ which evaluates to $-2\over\sqrt{3}$. The method in the book is to factor out $x^2$ from the root in the denominator:
$$\begin{align}
\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}} & = \lim_{x\to-\infty} {2x-1\over \sqrt{x^2\left(3+\frac{1}{x}+\frac{1}{x^2}\right)}} \\
 & = \lim_{x\to-\infty} {2x-1\over -x\sqrt{3+\frac{1}{x}+\frac{1}{x^2}}} \\
 & = \lim_{x\to-\infty} {-2+\frac{1}{x}\over \sqrt{3+\frac{1}{x}+\frac{1}{x^2}}} \\
 & = {-2\over\sqrt{3}}
\end{align}$$
the second step is justified because $x\to-\infty$ implies $x\lt0$, so $\sqrt{x^2}=-x$.
For my attempt I ended up with the negative of the correct answer:
$$\begin{align}
\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}} & = \lim_{x\to-\infty} \left({2x-1\over \sqrt{3x^2+x+1}}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\right) \\
 & = \lim_{x\to-\infty} {2-\frac{1}{x}\over \sqrt{\frac{1}{x^2}\left(3x^2+x+1\right)}} \\
 & = \lim_{x\to-\infty} {2-\frac{1}{x}\over \sqrt{3+\frac{1}{x}+\frac{1}{x^2}}} \\
 & = {2\over\sqrt{3}}
\end{align}$$
Where have I gone wrong? I suspect the mistake lies in my second step, but I'm unable to identify what went wrong exactly.
 [/INST] Your mistake is in writing
$$\frac 1 x = \sqrt{\frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
</s>