The derivative at a point $x$ is defined as: $\lim\limits_{h\to0} \frac{f(x+h) - f(x)}h$

But if $h\to0$, wouldn't that mean: $\frac{f(x+0) - f(x)}0 = \frac0{0}$ which is undefined?

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    $\begingroup$ You should imagine what happens as $ h $ gets closer and closer to $0$, not what happens when $ h=0$. $\endgroup$ – littleO Oct 23 '15 at 11:05
  • $\begingroup$ This is precisely why you learnt the theory of limits: to solve the indeterminate cases by extrapolating the value from nearby points. $\endgroup$ – Yves Daoust Oct 23 '15 at 13:44
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    $\begingroup$ You have assumed that $\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\frac{\displaystyle\lim_{h\to 0}(f(x+h)-f(x))}{\displaystyle\lim_{h\to 0}h}$. Why so? $\endgroup$ – user 170039 Oct 23 '15 at 14:01
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    $\begingroup$ Maybe it helps if you try to think about some very simple example, like $f(x)=x$. I.e., you get the limit $\lim\limits_{h\to0} \frac{(x+h)-x}h$. $\endgroup$ – Martin Sleziak Oct 23 '15 at 14:24
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    $\begingroup$ "Limits are about the journey, not the destination." $\endgroup$ – Blue Oct 24 '15 at 8:17

Instead of $h$ here I will use $\Delta x$ as I think it makes more sense to consider a change in $x$.

For $$\lim\limits_{\Delta x\to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}\tag{1}$$ at no point do we ever substitute $\Delta x=0$. We can only say that in the limit $\Delta x$ tends to zero; this is why there is the limit notation directly in-front of the fraction.

Consider the graph below which is an identical graphical representation of Equation $(1)$:

limit graph

You can see from the graph that the Secant line, which is the line that intersects the purple curve at $(x,f(x))$ and at $(x+\Delta x,f(x+\Delta x))$ is an approximation to the derivative (tangent) of the purple curve at $(x,f(x))$.

Now imagine $\Delta x$ getting smaller and smaller until eventually it becomes vanishingly small; which we call 'infinitesimal' (loosely speaking this is as small as you can possibly get but still greater than zero).

At this point you can see that the Secant line approaches the Tangent line or derivative at $(x,f(x))$ as $\Delta x$ tends to zero. Thus, the approximation of the Secant line to the Tangent line is optimal at $(x,f(x)+\Delta x)$ and this is the exact meaning of Equation $(1)$.


You seem to conflate the two notions of "take the limit as $h$ approaches $0$" and "plug in $0$ for $h$". In some situations, the latter is an easy way of doing the former. But in other situations, including the definition of derivative, the latter is (as you noticed) nonsense, and so you have to actually do the former. (The first step toward doing the former is, of course, to make sure you thoroughly understand the concept of limit.)


When I taught Calculus, I got this question a lot. One simple exercise that helped many students understand the concept of a limit is the following: Pick your favorite function, e.g. $\sin(x)$, then evaluate the fractional part of the limit with a calculator choosing some small $h$ and some value for $x$. For example with $x=1$ and $h=0.1$, we get: $${\sin(1.1) - \sin(1) \over{0.1} }= 0.497364$$ Then choose a smaller $h$, but keep $x$ the same. For example, for $h=0.01$: $${\sin(1.01) - \sin(1) \over{0.01} }= 0.536086$$ And continue with smaller and smaller values for $h$. \begin{array}{|c|c|} \hline h & (\sin(x+h) - \sin(x))/h \\\hline 0.1 & 0.497364 \\\hline 0.01 & 0.536086 \\\hline 0.001 & 0.539881 \\\hline 0.0001 & 0.540260 \\\hline 0.00001 & 0.540298 \\\hline 0.000001 & 0.540302 \\\hline \end{array} You can clearly see that the value is converging to the derivative which is: $$\cos(1) = 0.5403023059$$

  • $\begingroup$ My teacher had us play a game called closer to zero. He would start with a number, like 1, and call on someone for a number that was closer to zero. You might say 0.1, for instance, and then the next person might say 0.05, then 0.00007. Much like your table's h values, they just get closer to zero. He did it sort of as a game, almost sarcastically, and despite being silly, it was quite helpful for explaining a limit of zero. $\endgroup$ – Dan Oct 23 '15 at 22:51

Actually, the $ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $ is the value which $\frac{f(x+h) - f(x)}{h}$ approaches when you keep reducing $h$ as much as you can.

Here is how my teacher explained me the idea:

Think of drawing a tangent. It's a line touching a curve at only one point.

But how could that be possible? Two points determine a line.

The tangent is the limiting case of a secant where the two points close in on each other.

Link to Animation

Like he said, and you correctly pointed out, the expression is meaningless if $h=0$.

Here is another paradox that Zeno used:

Suppose an Arrow is shot which travels from A to B.

Consider any instant.

Since, no time elapses during the instant, the arrow does not move during the motion.

But the entire time of flight consists of instances alone.

Hence, the arrow must not have moved.

The key here is to introduce the idea of speed, which is once again, a limit, the distance travelled per duration as the duration approaches an instant (tends to zero).


The definition $\mathbf{strictly}$ means $h\to 0$ and NOT $h=0$. Thats where you are mistaken. So you can't put $h=0$ directly in the formula, rather you have to calculate the limit of that expression for a particular $f(x)$ as $h\to 0$.

  • $\begingroup$ But as h->0 wouldn't f(x+h) -> f(x) ? I had read that for continuous functions, the lim x->c f(x) = f(c) and for a function to be differentiable, doesn't it have to be continuous? Then why is f(x+h)-f(x)/h undefined when I plug in 0? $\endgroup$ – User2956 Oct 23 '15 at 10:50
  • $\begingroup$ You have answered your own question. As $h\to 0$ , $f(x+h)\to f(x)$ and not $f(x+h) = f(x)$. And yes for continuous functions, $\lim_{x\to c} f(x)=f(c)$. But that does not imply that $\frac{f(x)}{x}$ is continuous at 0 as in the defn. Check that $f(x)$ is differentiable at $c$ implies $f(x)$ is continuous at $c$ . But this has nothing to do with the existence of the limit $\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$. $\endgroup$ – SchrodingersCat Oct 23 '15 at 11:11

Think of how fast some terms of $f(x+h)-f(x)$ vanish compared to $h$ (no matter how small $h>0$ is), while others are independent of it.

In particular, $f(x+h)-f(x)$ consists of the derivative times $h$ (hence independent of $h$), plus a sublinear term $u(h)$ that tends faster to 0 than $h$, i.e. $\lim_{h \to 0} \frac{u(h)}{h} = 0$.

To illustrate this, take $f(x) = x^3$. Then $$f(x+h)-f(x) = x^3 + 3xh^2 + 3x^2h + h^3 - x^3 = 3x^2 h + h^2(3x + h)$$

The derivative term is $3x^2$, while $u(h) = h^2(3x+h)$ satisfying $\lim_{h \to 0} \frac{u(h)}{h} = 0$. Thus, $$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} 3x^2\frac{h}{h} + \lim_{h \to 0} \frac{u(h)}{h} = 3x^2 + 0$$


It might be helpful to consider the alternate notation in this case as written in: https://en.wikipedia.org/wiki/Derivative

Consider $\lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}$ where $y = f(x)$. Under this notation you'll note that the denominator is not a variable per say, but a difference. In your notation $h = \Delta x$. This notation can be misleading, leading you to conclude that $h$ is a variable that can be substituted. In fact (as eluded to by others) $\Delta x = 0$ can't be "reached", hence the need for a limit.


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