[Math] How to find the Horizontal and Vertical asymptotes of $\frac{x}{(x^4+1)^{\frac{1}{4}}}$

asymptoticscalculuslimits

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?

BTW

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

Best Answer

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

BOBO BOTS EATS DC

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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