when $f(x)^n$ is a degree of $n$ why is useful to think $\sqrt{f(x)^n}$ as $n/2$? I have come across this question when doing problems from "Schaum's 3000 Solved Calculus Problems".  I was trying to solve $$\lim_{x\rightarrow+\infty}\frac{4x-1}{\sqrt{x^2+2}}$$  and I couldn't so I looked the solution and solution said 
Can someone please explain to me why this is and exactly how it works? Also, the next question is as such $$\lim_{x\rightarrow-\infty}\frac{4x-1}{\sqrt{x^2+2}}$$ and there the author has suggested that $x= -\sqrt{x^2}$. Why is that? 
Thanks
EDIT
Can someone use the above technique and solve it, to show it works? Because I understand the exponent rules, I am aware of that but what I don't understand is why you want to do that?
Here is the solution that book shows:

 A: Note that $$-|x|-\sqrt{2}\leq\sqrt{x^2+2}\leq|x|+\sqrt{2}$$
From the last inequality, you can conclude that the rate of growth of the function $\sqrt{x^2+2}$, is in some sense linear or in the language of the author, the degree of $\sqrt{f(x)}$ is something like $\frac{n}{2}$. Therefore the functions $4x-1$ and $\sqrt{x^2+2}$ are in some sense proportional , or better saying, there exists the limit: $$\lim_{x\rightarrow\infty}\frac{4x-1}{\sqrt{x^2+2}}$$
A: As you may know $(\alpha^n)^m = \alpha^{n.m}$, since the action of raising something to the power of $n$ can be thought of as an inverse action of taking the $n$-th root.
So, we define $\sqrt[m]{a^n} = a^{\frac{n}{m}}$ (this is because, division is the inverse operator of multiplication), where $\frac{m}{n}$ is in lowest terms. And in fact, this definition is valid.
Examples
Some other examples:


*

*$\sqrt{2} = 2^{\frac{1}{2}}$

*$\sqrt[5]{a^3} = a^{\frac{3}{5}}$

*$\sqrt[4]{a^2} = a^{\frac{2}{4}}$, this is incorrect, as $\frac{2}{4}$ can be further reduced.

*$\sqrt[2]{a^2} = a^{\frac{2}{2}} = a$, this is wrong too, the same reason as above. It should read $\sqrt[2]{a^2} = |a|$ instead.


Applying it here, we have: $\sqrt{a^n} = a^{\frac{n}{2}}$, which means, if we take the square root of some $n$ degree term, we'll have a term of degree $\frac{n}{2}$.

Secondly, $sqrt{x^2}$ is always non-negative. Like this:


*

*$\sqrt{2^2} = \sqrt{4} = 2$

*$\sqrt{(-2)^2} = \sqrt{4} = 2$

*$\sqrt{3^2} = \sqrt{9} = 3$

*$\sqrt{(-3)^2} = \sqrt{9} = 3$


so, if x is nonnegative, we'll have $x = \sqrt{x^2}$, but when x is negative, then we'll have to put a minus sign in front of $\sqrt{x^2}$ to make it negative too, so $x = -\sqrt{x^2}$.
As x tends to $+\infty$, x will be positive, and vice versa, when x tends to $\infty$, it'll be negative.

Remember that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ as long as $a; b \ge 0$.
Example
Evaluate $\lim\limits_{x \rightarrow -\infty} \frac{-5x}{\sqrt{x^2+x}}$.
$\begin{align}\lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2+x}} &= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2 \left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2} \sqrt{\left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{-x \sqrt{\left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{5}{\sqrt{\left(1+\dfrac{1}{x} \right)}} \quad \mbox{cancel } -x\\
&= \dfrac{5}{\sqrt{\left(1+0 \right)}}\\
&= 5 \end{align}$
A: Take the maximum power of $x$ common in the numerator and the denominator. In this case, when you take $x^2$ common in the square root, it is the same as taking $x$ common in the numerator and denominator. Cancel $x$. The answer to the first question should be 4.
A: That is because if $f(x)$ has degree $2n$ then $\sqrt{f(x)}=O(x^n)$ when $x\to\infty$.
Elaborately, $\displaystyle\lim_{x\to\infty}\left|\frac{\sqrt{f(x)}}{x^n}\right|=M$, where $M\in\mathbb{R}$.
A: In these types of problem where both the numerator and denominator tend to $\infty$ we try to get rid of $\infty$ as much as possible, so we see if it is possible to cancel one $x^1$ or more $x^n$, for this reason treat a function under square root sign as function with degree $1/2$ if $\sqrt f(x)$ OR $n/2$ if $\sqrt f(x)^n$ and proceed with the cancellation procedure.
