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My textbook asks me to evaluate the limit $$lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}}$$ which evaluates to $-2oversqrt{3}$. The method in the book is to factor out $x^2$ from the root in the denominator:

$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} {2x-1over sqrt{x^2left(3+frac{1}{x}+frac{1}{x^2}right)}} \
& = lim_{xto-infty} {2x-1over -xsqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = lim_{xto-infty} {-2+frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {-2oversqrt{3}}
end{align}$$

the second step is justified because $xto-infty$ implies $xlt0$, so $sqrt{x^2}=-x$.

For my attempt I ended up with the negative of the correct answer:

$$begin{align}
lim_{xto-infty} {2x-1over sqrt{3x^2+x+1}} & = lim_{xto-infty} left({2x-1over sqrt{3x^2+x+1}}cdotfrac{frac{1}{x}}{frac{1}{x}}right) \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{frac{1}{x^2}left(3x^2+x+1right)}} \
& = lim_{xto-infty} {2-frac{1}{x}over sqrt{3+frac{1}{x}+frac{1}{x^2}}} \
& = {2oversqrt{3}}
end{align}$$

Where have I gone wrong? I suspect the mistake lies in my second step, but I'm unable to identify what went wrong exactly.

Your mistake is in writing

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

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