How do I find the horizontal and vertical asymptotes of the following?: $\frac{x}{(x^4+1)^{\frac{1}{4}}}$

Based on the definition of being a horizontal asymptote, I must therefore find out the limit as x approaches positive and negative infinity

But I tried to rationalize the denominator but in vain and I was wondering what would be the best method of carrying out this problem?


My school textbook stated that I must multiply the fraction with : $\frac{\frac{1}{x}}{\frac{1}{(x^4)^{\frac{1}{4}}}}$

But that confused me because the numerator and the denominator are not of the same value and thus wouldn't that be incorrect?

Please excuse my inadequecy of calculus

  • 1
    $\begingroup$ Are you asking if $(x^4)^\frac{1}{4}$ is $x$? $\endgroup$ – randomgirl Sep 6 '17 at 2:11
  • $\begingroup$ So were these answers helpful? $\endgroup$ – user334639 Oct 16 '17 at 17:59

I would suggest something slightly different from what the textbook suggested.

Start with the observation that $$ x = \begin{cases} (x^4)^{1/4}, & x>0 \\ -(x^4)^{1/4}, & x<0 . \end{cases} $$

Now you are ready to multiply both numerator and denominator by $(x^{-4})^{1/4}$, so you obtain $$ \lim_{x\to\pm\infty} \frac{x}{(x^4+1)^{1/4}} = \lim_{x\to\pm\infty} \frac{\pm (x^4)^{1/4}}{(x^4+1)^{1/4}} = \lim_{x\to\pm\infty} \frac{\pm (x^4)^{1/4}(x^{-4})^{1/4}}{(x^4+1)^{1/4}(x^{-4})^{1/4}} = \lim_{x\to\pm\infty} \frac{\pm 1}{(1+x^{-4})^{1/4}} = \pm \frac{1}{(1+0)^{1/4}} = \pm 1. $$

So the horizontal asymptotes are $y=1$ and $y=-1$.

There are no vertical asymptotes because the function is continuous at every $x\in\mathbb{R}$.


It’s always good to check for vertical asymptotes where the function is not defined (after you factor out removable discontinuities). The function $$\frac{x}{\left( x^4+1 \right)^{1/4}}$$ does not exist when we have a divide-by-zero situation; in other words, where

$$\begin{align} \left( x^4+1 \right)^{1/4} &= 0 \\ x^4+1 &= 0 \\ x^4 &= -1 \\ \end{align}$$

So it looks like we don’t have any vertical asymptotes on the real line—sweet!

Horizontal asymptotes are a limit as $x\to\pm\infty$. I’ll teach you a snazzy little acronym that we learned in algebra and used in calculus


For rational functions, if the power

  • IS Bigger On Bottom, the horizontal asymptote as $x\to\infty$ is $0$
  • is Bigger On Top, the asymptote is a Slant asymptote (irrelevant here)
  • if the Exponents Are The Same, you Divide Coefficients

EATS DC comes with a caveat: you just look at the leading power of $x$, the degree of the numerator and denominator polynomial jumbo.

So we know that the polynomial would look something like this if we solved the radical:

$$\frac{x}{ \sqrt[4]1 \cdot x^{4/4}+\cdots} = \frac{x}{1x^1+\cdots}$$

and that’s good enough! (Note: always be careful with the coefficients when taking care of that exponent.) Of course, $1/1=$, so we have

$$\text{horizontal asymptote: }y=1$$ $$\lim_{x\to\infty}\left[ \frac{x}{\left( x^4+1 \right)^{1/4}} \right]=1$$

But there’s a catch!

Because this function is symmetrical about the origin, it also holds that

$$\text{horizontal asymptote: }y=-1$$ $$\lim_{x\to-\infty}\left[ \frac{x}{\left( x^4+1 \right)^{1/4}} \right]=-1$$

This is because the $x$ up top introduces the negative sign as $x\to-\infty$ whereas the $1$-over-even-power exponent makes $\left( x^4+1 \right)^{1/4}$ positive for negative $x$.

Looking at a graph (which I assume you aren’t allowed to do) confirms all three of these! (I recommend Desmos.)


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