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.

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    $\begingroup$ Also, thank you very much for including a full explanation of your thoughts, enough context to know what you can use, and a clear identification of where you think your error is. This is a well-written question. $\endgroup$ – T. Bongers Dec 11 '18 at 2:46
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    $\begingroup$ It is worth pointing out that we must always check the conditions before applying any 'rule'. Relevant to this question is the fact that for reals $a,b,c$ we have $(a^b)^c = a^{b·c}$ if $a$ is positive, and not necessarily so otherwise: $((-1)^6)^{1/2} \ne (-1)^{6·1/2}$. $\endgroup$ – user21820 Dec 11 '18 at 9:08
  • $\begingroup$ In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis. $\endgroup$ – Gautam Shenoy Dec 11 '18 at 14:55
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    $\begingroup$ Note that $\displaystyle\,\sqrt{\,{a^{2}}\,}\, = \left\vert\,{a}\,\right\vert$ $\endgroup$ – Felix Marin Dec 12 '18 at 5:33

Your mistake is in writing

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

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


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