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.

Solution 1:

Your mistake is in writing

$$\frac 1 x = \sqrt{\frac{1}{x^2}}.$$

Since $x < 0$, the correct version includes a negative sign.