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 when a polynomial of degree n it is useful to think of the degree as being n/2

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?



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: enter image description here

  • $\begingroup$ Your subject says $f(x)^n$ and $\sqrt{f(x)^n}$ but the quoted text just says $f(x)$ and $\sqrt{f(x)}$. The text seems poorly edited - I suspect they meant to say "as being of degree $n/2$." $\endgroup$ – Thomas Andrews Feb 7 '13 at 14:49

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.


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 $a; b \ge 0$.


Evaluate $\lim\limits_{x \rightarrow -\infty} \frac{-5x}{\sqrt{x^2+x}}$.

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

  • $\begingroup$ Ahh I get it. Could you please also explain how this plays into solving the problem? $\endgroup$ – Jeel Shah Feb 7 '13 at 15:43
  • $\begingroup$ I've edited my post to give you another example, so that you can give your problem a try. You can notice a little bit further that, when I pull $x^2$ (order 2) out of the square root, it becomes $x$ (order 1). So a terms of order $n$ inside a square root is actually order $\frac{n}{2}$. $\endgroup$ – user49685 Feb 7 '13 at 16:00
  • $\begingroup$ Great! Thank you very much. Amazing answer :D $\endgroup$ – Jeel Shah Feb 7 '13 at 16:06
  • $\begingroup$ Could you please expand your answer before you have put $=5$ Like quickly explain the result of the denominator? $\endgroup$ – Jeel Shah Feb 7 '13 at 16:14
  • 1
    $\begingroup$ I cancel -x from both numerator, and denominator; then just kinda plug the numbers in. As x grows big, $\frac{1}{x}$ will tend to 0, you can take big value for x, like 1000000, or 1000000000, and see why. $\endgroup$ – user49685 Feb 7 '13 at 16:20

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: $$\lim_{x\rightarrow\infty}\frac{4x-1}{\sqrt{x^2+2}}$$


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.


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 $M\in\mathbb{R}$.

  • $\begingroup$ Is $O$ some function? $\endgroup$ – Jeel Shah Feb 7 '13 at 15:14
  • $\begingroup$ @gekkostate It is the big-oh notation. You can read a little about it with some googling. $\endgroup$ – Pedro Tamaroff Feb 7 '13 at 15:53

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.


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