I'm trying to follow an example in Kline's Calculus book. He introduces limits using an equation for speed and distance:

(1) $$s=16t^2$$

And the following speed deduced for some interval where h is not zero:

(2) $$\frac {k}{h} = \frac{128h+16h^2}{h}$$

But he says as $h$ tends to zero, the speed tends to 128, because the above when h is not zero, dividing the denominator and numerator by h gives:

(3) $$\frac {k}{h} =128+16h $$

And as $h $ tends to zero so does 16h.

But why does the function tend to 128 and not, say, infinity? (Since the denominator tends to zero?)

I have also been following MIT opencourseware 18.01 so have a deeper understanding than this point in the book but ran into the same issue unresolved. I have also covered calculus in other (formal) courses that I successfully completed, but this is a real stumbling block.

As a general principle for all functions at this level, why would the $k/h$ equation such as (2) not tend to infinity? Why is it legal to go from (2) to (3)? Why is the denominator (seemingly arbitrarily) not the dominant part? As is the case with any difference quotient as far as I'm aware?

I need an extremely convincing and intuitive answer as I have thought about this endlessly and have found no examples where this makes sense to me! I think I need an answer from someone who has previously very much struggled to understand this concept. I will examine very closely existing answers here, but I fear I will still need to drill down to find the most convincing (but indeed incorrect) argument for the paradox, as I still have not done a good job of pin pointing my confusion. I say this because I think to a degree I understand the responses so far, and even the above but my lack of confidence still remains. I think it would be better to consider an unknown continuous function $f (x) $ and the quotient:

$$\frac {\Delta f}{\Delta x} = \frac {f (x) - f(x_0)}{x - x_0}$$

even though I know $f$ needs to be substituted first before computing the limit.

Thank you and apologies to all trying to help!!

  • 3
    $\begingroup$ Both the denominator and numerator tends to zero; the question is how fast they tend to zero $\endgroup$
    – Lazy Lee
    May 24, 2017 at 23:41
  • $\begingroup$ Thanks but your comment is not helpful [for me!]!!! I need someone to address my level of stupidity and entrenched misunderstanding :-/ $\endgroup$
    – Jodes
    May 24, 2017 at 23:44
  • $\begingroup$ Shouldn't it be $dk/dh$? $\endgroup$
    – user301988
    May 24, 2017 at 23:44
  • $\begingroup$ @selfawareuser yes, or delta s and delta t with (2) being the difference quotient effectively, but he chooses to use k and h before launching into calculus proper by explaining limits in its own right $\endgroup$
    – Jodes
    May 24, 2017 at 23:46
  • $\begingroup$ For the benefit of those of us who don't have a copy of the book, how does Kline introduce the symbol $k$? It is clear from the equations that $k$ is not intended to be a constant, but what is it supposed to be? $\endgroup$
    – David K
    May 25, 2017 at 2:50

4 Answers 4


Although plugging in numbers can be misleading a lot of the time, I think it's helpful here.

What happens if we take some $h$ which is "close to $0$" - say, $h={1\over 1000}$? Then the fraction is $${0.128+.000016\over 0.001}={0.128016\over 0.001}=128.016.$$ Now try $h={1\over 10^4}$, $h={1\over 10^5}$, ...

You'll quickly observe that they seem to tend towards $128$. Now obviously that doesn't prove that the limit is $128$, but it suggests it.

The problem is that the instinct "small denominators yield small fractions" ignores the possibility of the numerator being small as well. In this case, as $h$ goes to zero, both the numerator and the denominator of the fraction go to zero, and the limit is going to wind up basically being a measure of how fast they go to zero relative to each other (and this can be made precise!).

OK, so now that we've dispensed with the false heuristic, what's the right answer?

In this case, the fraction is reasonably simple, and we can analyze it directly: $${128h+16h^2\over h}=128+16h,$$ and this clearly goes to $128$ as $h$ goes to zero. In other cases things can be more complicated (what's the limit as $h$ goes to zero of $\sin(h)\over h$?), but in this case it's straightforward.


The expression $\dfrac 5 0$ is undefined since there is no number $x$ for which $0x=5.$

But $\dfrac 0 0$ is different: There are infinitely many numbers $x$ for which $0x=0.$

If $f(x)$ approaches $5$ and $g(x)$ approaches $0$ then $f(x)/g(x)$ approaches $\infty$ (here I mean the $\text{“}\infty\text{''}$ that is approached if you go in either the positive or the negative direction, rather than $+\infty$ or $-\infty,$ but in some cases one can say the limit is $+\infty$ or $-\infty.$

If you ask how many millionths of a mile a car travels in one millionth of an hour, and divide the former near-zero quantity by the latter, you will get the average number of miles per hour the car goes during that millionth of a second. If you let that millionth of a second approach $0$, you will get the car's speed at a particular instant, in which it goes $0$ miles in $0$ hours, regardless of what its speed is at that time. That limit will not generally be $\infty.$

That's what derivatives are. They are limits as the numerator and denominator both approach $0.$ If that were always $\infty,$ then differential calculus would not exist.


Here is a quick internal consistency check of your logic.

$2=\frac {2n} n$

As $n \to 0$ would you say $2 \to \infty$, off course, you would not!

Num and Den going to zero is not a sufficient condition for the limit going to $\infty$ as a whole.

I have always found it useful to test my intuitions against such small tests.


The short answer is that these are limits , you are working with numbers near of $0$ but not equal to $0$

So if you have that a number $n$ tends to zero means that $n$ is near of $cero$, e.g. $10^{-99} = 0.\underbrace{000...}_{98 times}1$

If you have $\lim_{n\to 0}\frac{kn}{n}$, you need to know that $n$ is not zero, therefore it can be canceled by algebra rules, to get the limit equal to $k$

However, if you have $\lim_{n\to0}\frac{k}{n}$ with $k \neq 0$, since $n$ is smaller and near of $0$ the fraction is bigger and bigger, therefore you can note that there is no limit($+\infty$ or $-\infty$). This, geometrically is interpreted as a vertical asymptote.


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