I just started AP Calculus A this semester (last semester was Precalculus/Math Analysis), and my teacher assigned us homework in which I have to get the vertical and horizontal asymptotes of a given function. However, he failed to explain to us how to get the vertical asymptotes without using any precalculus. I'd like to avoid the precalculus and learn as much as I can early on, so I've been trying to do it with the little calculus I've been taught, but can't find any online resources. Given the below example, this is the farthest I've gotten:

For the horizontal asymptote...

$$\mathrm{evaluate:}\quad\lim_{x\to ∞}\,\left({x\over x+4}\right)$$

For the vertical asymptote...

$$\mathrm{solve:}\quad\lim_{x\to a}\,\left({x\over x+4}\right) = ∞$$

The first I can already solve, but it's the vertical asymptote where I have no idea what to do. I can't just substitute a in for x and try to solve ${a\over a+4}=∞$, because that's technically impossible.

How would I solve for a without graphing out the equation and visually identifying the vertical asymptotes?


To find the vertical asymptote, you don't need to take a limit. Instead, find where the function is undefined. For $f(x)=\frac{x}{x+4}$, we should find where $x+4=0$ since then the denominator would be $0$, which by definition is undefined.

Solving this, we find that a vertical asymptote exists at $x=-4$. We know this is an asymptote rather than a removable discontinuity because there is nothing we can do to manipulate the function to include $x=-4$ in the domain.

Intuitively, this means that at the vertical line $x=-4$, the function does not "exist". Instead, we approach $\pm \infty$, depending on which side we come in on.

  • $\begingroup$ This is what we were taught in class, but I wasn't sure if there was some more... I guess you could say "interesting" way to do it with limits. Thanks! $\endgroup$ – jmindel Jan 13 '16 at 4:42
  • $\begingroup$ Glad to help! What you could say is that $\lim \limits_{x \to -4^+} \frac{x}{x+4}=\infty$ and $\lim \limits_{x \to 4^-} \frac{x}{x+4}=-\infty$ $\endgroup$ – zz20s Jan 13 '16 at 4:43
  • $\begingroup$ What were you saying? Sorry, it cut off at the start of your MathJax. So is there really any way to solve for a either way, without already knowing to check as x approaches $-4$? $\endgroup$ – jmindel Jan 13 '16 at 4:44
  • $\begingroup$ I did not so much as solve it another way, I just formalized the "intuitive" definition of the asymptote using a limit. $\endgroup$ – zz20s Jan 13 '16 at 4:45
  • $\begingroup$ Ok. Thanks again! $\endgroup$ – jmindel Jan 13 '16 at 4:47

The limit of $\frac{g(x)}{f(x)}$ is $\infty$ when $g(x)\neq 0$ and $f(x)=0$. So just solve for when the denominator is $0$. This will give you $a=-4$.

  • $\begingroup$ As $x \to -4^{+}$, $x$ is negative and $x + 4$ is positive, so $$\lim_{x \to -4^{+}} \frac{x}{x + 4} = -\infty$$ As $x \to -4^{-}$, both $x$ and $x + 4$ are negative, so $$\lim_{x \to -4^{-}} \frac{x}{x + 4} = \infty$$ Since the left- and right-sided limits disagree, the limit does not exist at $x = -4$. $\endgroup$ – N. F. Taussig Jan 13 '16 at 13:03
  • $\begingroup$ You're completely right, I was being too loose with the limit of $\frac{g(x)}{f(x)}$. I should have just said there will be a vertical asymptote. $\endgroup$ – Tony S.F. Jan 13 '16 at 21:48

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