Can we find out vertical asymptotes by finding the limit of a function y=f(x)/g(x) when y approaches infinity?

Question

A vertical asymptotes occurs when y approaches infinity (like a horizontal asymptote occur when x approaches infinity). So, can you apply a limit as y approaches infinity, on any function, especially rational functions, to find the vertical asymptotes?

Why is it that vertical asymptotes are defined as the 'the values x can't equal'? How is the related to the limit as y approaches infinity?

Please consider that I am a high school student studying high level maths, so answer with explanations that you think I will understand.

The vertical asymptotes are not defined as "the values $x$ can't equal." For example, the function $f(x)=\frac{x(x+1)}{x+1}$ is not defined at $x=-1$, but $f(x)$ does not have a vertical asymptote there. Rather, a rational function has a vertical asymptote at $x=c$ if $\lim_{x\to c^{\pm}}f(x)=\pm\infty$. (That last expression means that $y$ tends to either $\infty$ or $-\infty$ as $x$ approaches $c$ from either the left or the right.) — symplectomorphic, Aug 21 '16 at 05:46
Also, now I'm confused as to why f(x) = x(x+1)/x+1 does not have an asymptote at x= -1. Yes you can remove x+1 and the function becomes linear. But if you substitute x = -1 like you have given the function, as above, then x cannot equal to 1. — Person, Aug 21 '16 at 05:50
Do you know what vertical asymptote means, visually speaking? It means the graph approaches a vertical line. If y=x(x+1)/(x+1) - notice the parentheses on the denominator by the way, don't forget that - is the same as y=x at every point except one, then it's pretty obvious from looking at the graph that it has no vertical asymptote. The rule that vertical asymptotes occur for x-values that cannot be plugged in only works when rational functions are expressed in simplest form (i.e. common factors between numerator and denominator cancelled out). — anon, Aug 21 '16 at 05:52
Yes, I've recognise asymptotes visually. So at x=-1, is there a discontinuity? — Person, Aug 21 '16 at 05:56
Got it, when graphing that, should I include the discontinuity? Most graphing calculators leave it out. — Person, Aug 21 '16 at 05:58

score 0 · Accepted Answer · answered Aug 21 '16 at 05:47

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Dividing by really small numbers yields really big numbers. Consider for instance the reciprocal function $1/x$. As you plug in numbers for $x$ that are closer and closer to $0$, the outputs of $1/x$ will get bigger and bigger in size (possibly negative, if you're plugging in negative numbers). This occurs with all rational functions $f(x)/g(x)$: as $x$ approaches a value in which $g(x)$ is $0$ but $f(x)$ isn't, you get division by smaller and smaller numbers yielding bigger and bigger outputs.

The $y$ coordinate of a point on the graph is the output of the function applied to the $x$ coordinate, so if the outputs are getting bigger in size, graphically that means the points on the graph are going way up or way down.

answered Aug 21 '16 at 05:47

anon

151,657

I see, now how do I apply this knowledge to finding a vertical asymptote. Do consider y = 4x^2 - 1/3x^2 - 5. How would I find the vertical asymptote for this? – Person Aug 21 '16 at 05:53
@Person Well, that's a polynomial and has no denominators, so there are no vertical asymptotes. Do you mean 4x^2-1/(3x^2)-5 instead? – anon Aug 21 '16 at 05:54
That will do fine. It just the limit that @symplectomorphic put in I want to understand through application or an example. – Person Aug 21 '16 at 05:57

score 0 · Answer 2 · answered Aug 21 '16 at 06:09

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If you can find the inverse of the function and hence write it in the form $x=f^{-1}(y)$ then you can find vertical asymptotes by finding the limits:

$$\lim_{y\to\infty}f^{-1}(y)$$

and

$$\lim_{y\to-\infty}f^{-1}(y)$$

This is exactly the same as what you are doing with $x$ to find horizontal asymptotes.

However depending on your function you may not be able to find an inverse function to do this with.

answered Aug 21 '16 at 06:09

Ian Miller

11,844

Well, why the inverse? And can you just let the denominator equal to zero, solve, and those are your asymptotes i.e. why are asymptotes where 'the values of x give y as undefined'? – Person Aug 21 '16 at 06:10
This will work for any function not just rational ones. Consider the simple case $y=\ln x$ then you have inverse $x=e^y$ and you find there is a asymptote $x=0$ because $lim_{y\to-\infty}e^y=0$. – Ian Miller Aug 21 '16 at 06:14
I see. Well if you take 2x^2 / x-4. x-4 = 0, x=4 is the vertical asymptote. How does this relate. And why is it that you must find the inverse? – Person Aug 21 '16 at 06:15
As for the why, think about what the inverse is - it basically involves swapping the $x$ and $y$. – Ian Miller Aug 21 '16 at 06:15
Ah, I see. So when you find the inverse, you basically do the same thing as when you find the limit of horizontal asymptotes. That makes sense. – Person Aug 21 '16 at 06:17
You asked "can you apply a limit as y approaches infinity". As others have pointed out there are other ways to find vertical asymptotes which do not involves limits so you do not have to find inverse unless you want to do it using limits. – Ian Miller Aug 21 '16 at 06:17
Yes. With the inverse you are doing basically the same thing. But in general its not going to be the easiest way to do it, especially for rational functions. – Ian Miller Aug 21 '16 at 06:19
Well the only other way I understand is through finding the values for which the denominator makes the function undefined and I have yet to understand why! – Person Aug 21 '16 at 06:19

Can we find out vertical asymptotes by finding the limit of a function y=f(x)/g(x) when y approaches infinity?

2 Answers2