Precalculus limits as x approaches negative infinity Example 1 $\displaystyle \lim _{x\rightarrow -\infty }\frac{\sqrt{9x^{2} +2}}{2x-9}$
Example 2
$\displaystyle \lim _{x\rightarrow -\infty }\frac{6x^{2} -x}{\sqrt{9x^{4} +7x^{3}}}$
I'm having difficulty understanding why in Example 1, we would divide the numerator and denominator by $-\sqrt{x^2}$, but in Example 2, we would divide the numerator and denominator by $+\sqrt{x^4}$. I know it has something to do with whether or not the exponent is even or odd, but I just need a little more clarification to get there. Thanks in advance.
 A: It would be far more convenient if, in the limit, almost every term were zero except for the "most important" term.  For polynomials, the most important term is the leading term (highest power of $x$), so let's factor that one out of the two polynomials we can see.  Recall that $\sqrt{a^2} = |a|$ (because the square root function cannot produce negative numbers) and as $x \rightarrow -\infty$, $|x| = -x$.
\begin{align*}
\lim _{x\rightarrow -\infty } \frac{\sqrt{9x^{2} +2}}{2x-9} 
    &= \lim _{x\rightarrow -\infty } \frac{\sqrt{9x^{2}(1 + \frac{2}{9 x^2})}}{2x(1-\frac{9}{2x})}  \\
    &= \lim _{x\rightarrow -\infty } \frac{|3x|\sqrt{1 + \frac{2}{9 x^2}}}{2x(1-\frac{9}{2x})}  \\ 
    &= \lim _{x\rightarrow -\infty } \frac{-3x\sqrt{1 + \frac{2}{9 x^2}}}{2x(1-\frac{9}{2x})}  \\
    &= \lim _{x\rightarrow -\infty } \frac{-3\sqrt{1 + \frac{2}{9 x^2}}}{2(1-\frac{9}{2x})}  \\
    &= \frac{-3\sqrt{1 + 0}}{2(1+0)}  \\
    &= \frac{-3}{2}  \text{.}
\end{align*}
Notice that what we have done is equivalent to dividing by $-\sqrt{x^2}$ : the minus sign because $|x| = -x$ and the $\sqrt{x^2}$ to represent that the polynomial under the radical has degree $2$.
If you are ever working mechanically and happen to factor out a non-leading term, you will discover this in the step where you evaluate the limit at every term.  You should only get one nonzero term in each of the numerator and denominator, but you will find that (at least) the leading term goes to infinity in this step.  So back up and factor out the correct leading term.
In your second example, recall that $3x^2 \geq 0$ for all $x$, so $|3x^2| = 3x^2$ for any $x$, even those approaching $-\infty$.
\begin{align*}
\lim_{x\rightarrow -\infty} \frac{6x^{2} -x}{\sqrt{9x^{4} +7x^{3}}}
    &= \lim_{x\rightarrow -\infty} \frac{6x^{2}(1 -\frac{x}{6x^2})}{\sqrt{9x^{4}(1 + \frac{7x^{3}}{9x^4})}}  \\
    &= \lim_{x\rightarrow -\infty} \frac{6x^{2}(1 -\frac{x}{6x^2})}{|3x^2| \sqrt{1 + \frac{7x^{3}}{9x^4}}}  \\
    &= \lim_{x\rightarrow -\infty} \frac{6x^{2}(1 -\frac{x}{6x^2})}{3x^2 \sqrt{1 + \frac{7x^{3}}{9x^4}}}  \\
    &= \lim_{x\rightarrow -\infty} \frac{2(1 -\frac{x}{6x^2})}{\sqrt{1 + \frac{7x^{3}}{9x^4}}}  \\
    &= \frac{2(1 -0)}{\sqrt{1 - 0}}  \\
    &= 2  \text{.}
\end{align*}
What we have done is equivalent to dividing by $+\sqrt{x^4}$: the plus sign because $|x^2| = x^2$ and $\sqrt{x^4}$ to represent that the polynomial under the radical has degree $4$.
Notice that, really, once you have factored the leading terms and (if needed) migrated them outside the radicals (taking care about signs), you are interested in how much those terms cancel and what single power of $x$ remains.  That power could be positive (larger degree in the numerator), zero (same degrees), or negative (larger degree in the denominator).  The remaining power tells you what the function does in the limit and the "bunch of constants" that were left behind may scale that behaviour by a constant multiple.
A: Hint:
Transferring the limit $\to0$ or $0^+$ makes calculation much easier
For the first case $$\lim_{x\to-\infty}\sqrt{9x^2+2}=\lim_{h\to0^+}\sqrt{\dfrac{9+2h^2}{h^2}}=\lim_{h\to0^+}\dfrac{\sqrt{9+2h^2}}{\sqrt{h^2}}$$
Now for real $h,\sqrt{h^2}=|h|$ and $|h|=+h$ for $h\ge0$
For the second case,  $$\lim_{x\to-\infty}\sqrt{9x^4+7x^3}=\lim_{h\to0^+}\sqrt{\dfrac{9-7h}{h^4}}=\lim_{h\to0^+}\dfrac{\sqrt{9-7h}}{\sqrt{h^4}}$$
