# 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.

• 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. – T. Bongers Dec 11 '18 at 2:46
• 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: $((-1)^6)^{1/2} \ne (-1)^{6·1/2}$. – user21820 Dec 11 '18 at 9:08
• In the future, it may help to substitute $x$ with $-x$ when dealing with limits on the negative axis. – Gautam Shenoy Dec 11 '18 at 14:55
• Note that $\displaystyle\,\sqrt{\,{a^{2}}\,}\, = \left\vert\,{a}\,\right\vert$ – Felix Marin Dec 12 '18 at 5:33

$$\frac 1 x = \sqrt{\frac{1}{x^2}}.$$
Since $$x < 0$$, the correct version includes a negative sign.