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:


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 $\lim\limits_{x \rightarrow -\infty} \frac{-5x}{\sqrt{x^2+x}}$.
$\begin{align}\lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2+x}} &= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2 \left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{\sqrt{x^2} \sqrt{\left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{-5x}{-x \sqrt{\left(1+\dfrac{1}{x} \right)}}\\
&= \lim\limits_{x \rightarrow -\infty} \dfrac{5}{\sqrt{\left(1+\dfrac{1}{x} \right)}} \quad \mbox{cancel } -x\\
&= \dfrac{5}{\sqrt{\left(1+0 \right)}}\\
&= 5 \end{align}$
