I've playing with the limit definition of derivative and I've to somewhat confusing conclusions.

To clarify, I'm from an Engineering background so I don't think that an instantaneous rate of change makes any sense, for it to be a rate of change something has to change, right?! So I think of derivatives as the ratio between a very small change in $y$ to a very small change in $x$. So It must be different at every point for many curves.

So I plugged in the formula to calculate the "ratio of change" between points $x+h$ and $x$

$\lim _{h\to \:0}\left(\frac{\left(x+h\right)^2-x^2}{x}\right) = 2x$

This is the expected answer

However If I try to plug in the points $x+6h$ and $x+5h$, both are distinct from $x+h$ and $x$, so at I expect to find a different answer but I found the same answer

$\lim _{h\to \:0}\left(\frac{\left(x+6h\right)^2-\left(x+5h\right)^2}{x}\right) = 2x$

It gets even more confusing If I try to compute the derivative between $x$ and a previous point $x-h$, it still gives me the same answer!

$\lim _{h\to \:0}\left(\frac{\left(x\right)^2-\left(x-h\right)^2}{x}\right) = 2x$

I can't make sense of it. It means If chose an arbitrary point $x = m$ then no matter how far I go far from it in any direction I get the same derivative which is quite patently wrong! If it is due to $h$ approaching zero, or it being an infinitesimal how then can we proceed from point to point if not by adding an infinitesimally lengthy line segment, or aren't all curves made up very small line segments?

I hope I made myself clear!

Thanks in advance!

  • $\begingroup$ You’ve got a consistent error in all of your difference quotients: the denominator should be $h$. $\endgroup$
    – amd
    Commented Jan 19, 2019 at 1:01
  • $\begingroup$ I wanted to write h. My bad, my bad! $\endgroup$ Commented Jan 19, 2019 at 9:36

2 Answers 2


Let's imagine the curve of an arbitrary function $f$. We can say that two points on the curve are $(x_1, f(x_1))$ and $(x_2, f(x_2))$

We can draw a straight line between these two points, and it's gradient is given by:

$$m=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ for any $x_1, x_2 \in \mathcal D(f)$ (the domain, essentially, wherever $f$ is defined)

We can assume that $x_2 > x_1$, or rather that $x_2 =x_1 +h$ for some arbitrary $h$ (In the case where it isn't, we can just switch $x_1$ and $x_2$ everywhere I've written them)

Then we have:


If $f(x)=x^2$, we have $m=\frac{(x_1+h)^2-x_1^2}{h}=\frac{h^2+2hx_1}{h}=2x_1+h$

What the derivative calculates is $m$ when $h$ is infinitesimal, that is, it measures the gradient between two points that are so close to each other they are basically the same point, and that forms the gradient of the tangent to $f$ at that point.

We can say that $$\lim_{h\to 0}{(2x_1+h)}=2x_1$$

In fact:

$$\lim_{h\to0} (kh)=0$$ is true for any constant $k$

and that is why the derivative always comes out as $2x$, because the $h$, and its multiples thereof, all disappear as the limit is taken to $0$.

  • $\begingroup$ But isn't it quite nonsensical to talk about a gradient at a point? What does it mean? $\endgroup$ Commented Jan 18, 2019 at 21:58
  • $\begingroup$ The gradient of the tangent to the curve at that point. $\endgroup$ Commented Jan 18, 2019 at 22:13
  • $\begingroup$ But how then the Gradient of the tangent to the curve tells us about change ALONG the curve? Don't we add these small changes to get from one point to another? Such as we do with integration? $\endgroup$ Commented Jan 18, 2019 at 22:28
  • $\begingroup$ For the first question, it tells us where the curve is about to go. For example, a negative derivative tells us the curve is about to decrease, and the quantity of that negative describes the scale (e.g. a derivative of $-5$ implies a much steeper descent than one of, say, $-2$). $\endgroup$ Commented Jan 19, 2019 at 3:38
  • $\begingroup$ For the second, no, you just calculate it from a different point. The reason why $2x$ was the derivative in all three cases was because $h \to 0 \implies kh \to 0 \forall k\in \Bbb R$. You were just adding multiples of $h$ in your other examples, hence they disappeared. With regards to integration, that finds the area under the curve between two points, and our approximation method for this, like the gradient method I described in my answer is an approximation for differentiation, is called the Trapezoidal Rule. $\endgroup$ Commented Jan 19, 2019 at 3:44

This is correct—you're just approaching $x$ in different ways as you take the limit. In the example with $f(x+6h)-f(x+5h)$ you're approximating the slope a little to the right of $x$, but as $h\rightarrow 0$ you still end up at $x$ and get the slope there—which turns out to be $2x$ as it should. You're deriving the same correct answer in different ways.

  • $\begingroup$ How can they be correct? How am I approaching the same point x? Then how can I define another point? How can I go from X0 to X1? $\endgroup$ Commented Jan 18, 2019 at 22:00
  • $\begingroup$ Isn't the interpretation of derivative is the rate of change between x and infinitesimally close point? Shouldn't they be different for each set if two points? What is the slope at a point? What is changing at a point which is a static entity? $\endgroup$ Commented Jan 18, 2019 at 22:04
  • $\begingroup$ @RaafatAbualazm It's nowadays very carefully not defined in terms of infinitesimal distances but as a limit—since a point can't have a slope. Geometrically, it's the slope of a tangent to the curve at the point. Another way to imagine it is that it's the direction you'd be facing if you were travelling along the curve—which you can define at each instant. $\endgroup$
    – timtfj
    Commented Jan 18, 2019 at 22:18
  • $\begingroup$ @RaafatAbualazm You're approaching $x$ each time because $\lim_{h\rightarrow 0} (x+5h) =x$ just like $\lim_{h\rightarrow 0} (x+h)=x$. $\endgroup$
    – timtfj
    Commented Jan 18, 2019 at 22:21
  • $\begingroup$ Then how can we reach point b from point a? Don't we add small changes to go from a to b? If that change is not along the curve of the function then how can the above reasoning and Integration hold? $\endgroup$ Commented Jan 18, 2019 at 22:26

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