# Why is derivative a two-sided limit?

Well, I haven't thought that I'll ask these kind of questions but well, really I do not really understand that.

Suppose we are making some secant lines in order to get the average rate of change and therefore we have points $x$ and $a$ - points of intersection with the function. Point $x$ is the point where we want to find an instantaneous rate of change. So what we are doing is making the second point closer and closer to $x$ and here is the problem.

We can do that from either left or right side. The second point can be a little smaller or bigger. We can take limit from the left or from the right ($+$/$-$). And as I know either way I will get the derivative. But actually the definition of the derivative is a TWO sided limit. So as we know from the limit properties in order to this to exist both the left and the right limits must exist and be equal. Why is the definition created like that even if I could find the derivative without bothering about all of these requirements? Do I need that constraint about the limits being equal? Why?

I am sure there are some reasonable arguments about why is that so even if I am not fully aware of them.

• It's a two-sided limit because the result should be independent of the direction of approach. If you don't require it being a two-sided limit, then the result is a one-sided derivative $f'(x+)$ or $f'(x-)$. – MPW May 15 '18 at 20:09
• well, what is derivative if the limits from either side are unequal? What would it represent? We do have definition for left and right derivatives for when the limits are unequal and the are clear that they refer left and right limits. This seems adequate to me. If we refer to a limit or a derivative the implication is that the left and right limits are equal. And it seems obvious that we can't have a meaningful definition if they are not. – fleablood May 15 '18 at 20:11
• "And as I know either way I will get the derivative." You know absolutely no such thing at all. Take $f(x) =|x^2- 9|$. As you approach from the left of $x = -3$ you get a limit of $2x$. If you approach from the right at $x = -3$ you get $-2x$. – fleablood May 15 '18 at 20:13

There are certainly examples where we could take the "derivative" from the left and right and get two different answers. For example, consider the absolute value function $f(x) = |x|$. As we approach $x=0$ from the right, all the secant lines have slope $1$; as we approach $x=0$ from the left, all the secant lines have slope $-1$.

In such an example, there is no well-defined tangent line (think about the graph of $|x|$). So we're interested in defining the derivative only where there is a well-defined tangent line, which is when the limits from the left and right of the secant slopes will be equal.

• Seems reasonable. But what about a situation when we have for example an existing completely normal right limit and not existing left. Does the derivative exist at that point? I mean by the definition clearly not, but why does the definition exclude this case? I can think of a "instantaneous rate of change" at that point in some sense. – Darono May 15 '18 at 20:59
• @Darono Such a rate of change would make sense considering change to the right, but not change from the left; it would only really make sense as a one- sided derivative. – BallBoy May 15 '18 at 21:16
• Answering Darono: Graph the function that is $f(x)=x$ for negative $x$ and $f(x)=x+1$ for non-negative $x$. The right derivative at $x=0$ exists and equals $1$, but the left derivative doesn’t exist, because the definitional limit goes to infinity. The definition prevents us from saying $f'(0)=1$, which is what it should do. – Steve Kass May 15 '18 at 21:27

Be reassured, the two-sided, left-only and right-only derivatives are all defined.

When both the left and right derivatives exist but are unequal, you have an angular point, which is an exceptional situation, where smoothness is broken.

It's to disqualify corners and cusps from being differentiable whereby the left derivative is different from the right derivative.

The fact is, for most nice continuous functions (polynomials, trig, exponential,...), the left/right derivates tend to equal each-other.So this is a pretty good justification for the definition.

For example $f(x+h) \approx f(x)+f'(x)h$ would otherwise only hold for $h>0$ if you defined the derivative as the positive limit. For such local approximations, we tend to care about what $f$ looks like both for $h>0$ and $h<0$, i.e. locally.

You seem to be making some circular thinking. You say “we can approach from either left or right, and get the derivative. So if we get the derivative when approaching from either direction, why is the derivative defined as a two-sided limit?”. Well, the fact that (for a differentiable function) you get the derivative either way is a result of the definition we have for the derivative. If it weren’t defined as a two-sided limit, you would not get the same value either way (because with a one-sided definition, you will have a lot of new “differentiable” functions that do not satisfy the original requirement).

I hope this makes at least as much sense as your question.

"We can take limit from the left or from the right (+ / - ). And as I know either way I will get the derivative."

You know absolutely no such thing at all.

Consider $f(x) = |x^2 -9|$ what is it's derivative at $x = -3$ and $x = 3$. You get two entirely different answers if you take them from the left or from the right.

The very short, non-responsive answer is that a limit is said to exist if both one-sided limits exist and agree. Since the derivative is a limit, its existence requires the existence and agreement of both one-sided limits.

Notice that all functions differentiable (in the usual sense) at $x_0$ are continuous and defined on an open set containing $x_0$.

There are six ways we could take limits of the difference quotient between $x_1$ and $x_2$, consistently choosing points on one side, at $x_0$, or on the other side, to get a derivative:

• $x_1 < x_2 < x_0$ and $x_1 \rightarrow x_0$, "left-left one-sided derivative",
• $x_1 < x_2 = x_0$ and $x_1 \rightarrow x_0$, "left one-sided derivative",
• $x_1 < x_0 < x_2$ and $x_1, x_2 \rightarrow x_0$ independently, "two-sided derivative",
• $x_1 < x_0 < x_2$ and $|x_1 - x_0| = |x_2 - x_0| \rightarrow 0$, "symmetric two-sided derivative",
• $x_0 = x_1 < x_2$ and $x_2 \rightarrow x_0$, "right one-sided derivative", and
• $x_0 < x_1 < x_2$ and $x_2 \rightarrow x_0$, "right-right one-sided derivative".

Notice, that the usual definition of derivative uses the left and right one-sided derivatives and requires that they agree.

If a function (exists on an open neighborhood of $x_0$, and is continuous on that neighborhood, and) is differentiable at $x_0$, all of these ways of computing a derivative agree. That is, if the usual derivative exists, any of these can be used to compute it. If the usual derivative does not exist, we can make pretty much anything happen, usually with piecewise functions. Examples:

Consider $f(x) = \begin{cases} -1 &,x<0 \\ 1 &, x \geq 0 \end{cases}$ and let $x_0 = 0$. Then $f(x)$

• is defined on $\mathbb{R}$,
• is discontinuous at $x_0$,
• has a left-left one-sided derivative of $0$ at $x_0$,
• has no left one-sided derivative at $x_0$,
• has no two-sided derivative at $x_0$,
• has no symmetric two-sided derivative at $x_0$,
• has a right one-sided derivative of $0$ at $x_0$, and
• has a right-right one-sided derivative of $0$ at $x_0$.

Consider $g(x) = |x|$. Then $g$

• is defined on $\mathbb{R}$,
• is continuous on $\mathbb{R}$,
• has a left-left one-sided derivative of $-1$ at $x_0$,
• has a left one-sided derivative of $-1$ at $x_0$,
• has no two-sided derivative at $x_0$,
• has a symmetric two-sided derivative of $0$ at $x_0$,
• has a right one-sided derivative of $1$ at $x_0$, and
• has a right-right one-sided derivative of $1$ at $x_0$.