I have been solving some exercises in Apostol, where he proves the asymptotes of the hyperbola. And I got the following question. When proving that the hyperbola approaches the asymptotes I used the definition of the equivalence:

$$\lim\limits_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 1$$

The intuition tells me that since $\exists r$, s.t. $\lvert \frac{f(x)}{g(x)} - 1 \rvert < \epsilon, \forall x>r$, the functions are essentially equal.

However, Apostol shows the result differently, using the limit of the difference:

$$\lim\limits_{x \rightarrow \infty} f(x) - g(x) = 0$$

What is the difference between the two approaches? I tried to prove the equivalence of statements, but I could not transform $\lvert \frac{f(x)}{g(x)} - 1 \rvert < \epsilon$ to $\lvert f(x)- g(x) \rvert < \epsilon$ easily.

Can someone show this equivalance, or tell me what is wrong? My intuition tells me that there is no difference between approaches, and pretty much each one shows that two functions are the same as x grows.

If these are different statements, why Apostol chose the latter approach?

  • 1
    $\begingroup$ Let $f(x) = x+1$ and $g(x) = x$. Then $\frac{f(x)}{g(x)} \to 1$ as $x\to \infty$, but $f(x) - g(x) = 1$ for all $x$. $\endgroup$ Jan 28, 2020 at 19:08
  • $\begingroup$ @Matthew, thx, nice counter example. So, does it mean the limit of difference is a stronger statement? Or they say different things overall? $\endgroup$
    – John
    Jan 28, 2020 at 19:13
  • $\begingroup$ @MatthewLeingang, in particular I m interested why it is necessary to show the asymptotes of hyperbola using the author's approach, instead of using the limit of quotient? $\endgroup$
    – John
    Jan 28, 2020 at 19:14
  • $\begingroup$ OK, I better answer the question instead. $\endgroup$ Jan 28, 2020 at 19:25

3 Answers 3


This is the difference between relative error and absolute error. Apostol shows that the absolute error between using one function to approximate the other is small, for instance, is eventually always less than $1$. You have shown that the relative error in doing so is small, for instance, is eventually always less than $1\%$ of the value of the function. The two are not equivalent in general.

Let's see how these compare in two scenarios.

  • We wish to approximate the function $x \mapsto x$ asymptotically as $x \rightarrow \infty$. The function $x+1$ has constant absolute error, $1$, and relative error that decreases as $1/x$. From this we can conclude that (absent additional hypotheses) relative error that decreases to zero does not imply absolute error decreases to zero.
  • We wish to approximate the function $x \mapsto 0$ asymptotically as $x \rightarrow \infty$. The function $1/x$ has absolute error decreasing to zero and relative error that is always undefined (division by zero). (If we replace $x \mapsto 0$ with $x \mapsto \varepsilon$ for some tiny, positive number, $\varepsilon$, the relative error is $|1 - \frac{1}{\varepsilon x}|$, which decreases for $x \in [1,1/\varepsilon]$ and increases afterwards, approaching $1$. The absolute error is bounded by $\varepsilon$.) From this, absolute error decreasing to zero does not imply relative error decreases to zero.

So the implication you are wanting between the two methods of measuring error doesn't exist in either direction, without more information about the various functions.

The problem with relative error is that it doesn't have to decrease for rapidly growing functions. Consider $2^x + x$. Its growth compared to $2^x$ is $$ \frac{2^x + x}{2^x} = 1 + \frac{x}{2^x} \xrightarrow{x \rightarrow \infty} 1 $$ but $$ (2^x + x) - (2^x) = x \xrightarrow{x \rightarrow \infty} \infty \text{.} $$ The error is small compared to the size of the function, so could be a negligible correction compared to the size of the function, but the error is eventually larger than any pre-specified bound on "small".


The conditions $\lim_{x\to \infty} \frac{f(x)}{g(x)} = 1$ and $\lim_{x\to\infty} (f(x) - g(x)) = 0$ are independent from each other:

  • [from my comment] Let $f(x) = x+1$ and $g(x) = x$. then $\frac{f(x)}{g(x)} \to 1$ as $x \to\infty$. but $f(x) - g(x) = 1$ for all $x$.
  • Let $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{2x}$. Then $f(x) - g(x) \to 0$ as $x \to \infty$, but $\frac{f(x)}{g(x)} = 2$ for all $x$.

As for why Apostol prefers the limit of the difference, consider the heuristic idea of asymptote: a line the curve becomes arbitrarily close to. The “closeness” is measured in terms of distance in the plane, so we should be looking at the difference $|f(x)-g(x)|$. Whereas, if you used the quotient definition, then every line $y= x+ c$ would fit the definition of “asymptote” for $y=x + \frac{1}{x}$.


A curve is asymptotic to a line if the distance between the line and the curve shrinks to zero when the variable that parametrizes the curve goes to infinity. So to show that your approach does not guarantee this behaviour it is enough to find a counterexample. Consider

$f(x) + x^2,\, g(x)=x^2+1$

Note that the distance between both curves remains finite, while it holds that

$\underset{x\rightarrow \infty}{\lim} \left | \frac{f(x)}{g(x)} \right | = 1$

The alternative approach, however, does not have this problem.

In essence, your approach measures the 'relative error' (the difference normalized by the size of the functions), while the notion of asymptote requires that the distance between the functions goes to zero in an absolute sense.


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