# Question about the limit definitions of derivative and definite integral

Actually this question may look simple and basic, but it is about something which bugs me for a long time, since when I took my first calculus classes. The limit $\lim_{x \to a}f(x) = L$ is defined when for each $\epsilon >0$ it is always $|f(x)-L|<\epsilon$ for some $\delta$ neighborhood of $a$, $0 < |x-a| < \delta$. This means that it doesn't matter how small $\epsilon$ gets, we are always able to find $f(x)$ values around $L$ in the neighborhood of $a$ within a $\delta$ distance.

In the definition of limit, the point $a$ is not needed to be included in the domain of $f(x)$. This is the part which I always find strange, when it comes to the definition of "solid" values which actually exist; but only at the "limit" and cannot be evaluated directly. For example, the derivative $f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$. Assume that this limit exists. Here $g(h) =\dfrac{f(x+h)-f(x)}{h}$ is a function of $h$ and it is not defined when $h=0$. The most instinctive thing would be to set $h=0$ and directly get derivative's exact value. But we can never directly put $h=0$ into $g(h)$ and find the exact value of the derivative. Despite this, the derivative clearly exists at $x$ and $h=0$ as a actual value, in the shape of the slope of the tangent line, passing through $(x,f(x))$. In this sense, I interpret here the limit operation as some kind of inference mechanism which fills in the "gap" at $g(h=0)$ by using the following logic : "We cannot exactly evaluate the function $g(h)$ at $h=0$ but we observe that there is a $L$ value such that for each $\epsilon > 0$, we can give a $\delta$ value such that all $g(h)$ values corresponding to $0 < |h| < \delta$ are contained between $L-\epsilon$ and $L+\epsilon$. The function cannot be evaluated as $g(0)=L$ but since all "nearby" values tend to get closer and closer to $L$ constantly, without any interruption, $L=f'(x)$ is the limit and can be used as a placeholder value for undefined $g(h=0)$". The placeholder is the critical word for me here: The derivative's exact value cannot be computed directly at $h=0$ since it results with a division by zero, but since the limit exists at $h=0$, it is considered as the most appropriate value, because all $h$ close to $0$ results in close $g(h)$ values s.t. $|g(h)-L| < \epsilon$.

I apply the same thinking to definite integral as well. Let the Riemann Sum between $a$ and $b$ for a function $f(x)$ is defined like that: $S(\delta x) = \sum_{i=1}^{n}f(x_i)\delta x + Excess$, where $n = \lfloor \frac{b-a}{\delta x} \rfloor$ and $Excess$ is the last partial rectangle in case $\delta x$ does not evenly divide $b-a$. Again, the exact area under $f(x)$ between $a$ and $b$ would be given if would be possible to evaluate $S(\delta x=0)$ but this is not defined, so we evaluate $\lim_{\delta x \to 0^{+}} S(\delta x)=A$ and if this limit exists, we deduce that the exact area under $f(x)$, between $a$ and $b$ would be the value of this limit $A$, since it would be the most natural value for $S(\delta x=0)$ if it were defined at that point.

My question is, is my thinking here true? Can we interpret the limit definitions of derivative and definite integral like that, assuming the limit at zero as the "most natural" value for them, since we cannot compute them analytically at $h=0$ and $\delta x=0$?

Yes. What you call the "most natural" value is the value that ensures continuity (when possible).

A function is continuous at a point when it has no "jump" there; this is formalized by the $\epsilon/\delta$ neighborhoods approach. When a function is undefined at a point, but nearby values are available, it is a natural choice to assign the value that ensures continuity. This is the very definition of a limit: we expect $f(x)$ to be the same as $f(x\pm\delta)$ with a small $\epsilon$ error.

Take some function defined as $f(x)=2+x$ for $x>0$ and undefined otherwise. If you want to extend the definition to $x=0$, the obvious choice will be $f(0)=2$. This extension is possible because you know the values of the function for points arbitrarily close to $0$. On the opposite, an extension of $f$ at $-1$ is not well defined.

This is the exact situation you face when defining the derivative or integral: they can only be defined in terms of limits and are the values that respect continuity. For example, let $f(h)$ be the slope of the chord of the parabola $y=x^2$ between the points $(1,1)$ and $(1+h,(1+h)^2)$: it equals $\frac{\Delta y}{\Delta x}=\frac{1+2h+h^2-1}{1+h-1}=2+h$, for $h>0$. But for $h=0$, this slope is undefined (we are missing a second point to define the straight line). If we are interested in the slope of the tangent, then we use continuity and admit that $f(0)=2$.

• Yes, this is just what I tried to describe in my question: Since we cannot evaluate derivative and integrals exactly, we follow the continuity of the functions which define them and deduce that the value which preserves the "flow of continuity" becomes our best choice for our actual derivative and integral values (or any other possible value which cannot be evaluated directly) and this value happens to be the limit. Is that right? – Ufuk Can Bicici Sep 3 '14 at 21:06
• Absolutely, the limit is a natural choice. The same holds for the "neighborhoods of infinity" and allows to sum infinite convergent series. – Yves Daoust Sep 3 '14 at 21:22
• By the way, since we define the derivative (tangent slope) and the definite integral (area) as limits which ensure continuity of some functions which approximate them, then these definitions are some kind of "axiomatic" things, is that right? Because there aren't any direct procedures which give their exact values, but we "intuitively" and "instictively" know that they are the limit values of some approximating "slope" or "area" functions? – Ufuk Can Bicici Sep 4 '14 at 6:39
• Yes. But it turns out by some magic that the properties of these items defined the analytic way coincide with equivalent properties defined by geometry. For example, the perpendicular to the tangent to a circle runs through the circle center. – Yves Daoust Sep 4 '14 at 7:27

Your intuitive description in the question is correct. The idea of limit is designed to handle situation where a function does not a value (because of the way it is defined). We think of the next best possibility that the function is defined around that troublesome point and then try to analyze the behavior of the function near the troublesome point.

If we are lucky, the function may behave very nicely near that point in the sense that its values remain near to some constant $L$ and then we say that the function has a limit at that point. Now it is up to our choice to define the value of function to be $L$ at that point (thereby making the function continuous at that point and getting rid of our issues regarding value of the function at a troublesome point). However we might not always be lucky and the limit of the function may not exist. In that case we have to live with the weird behavior of the function near that point.

However it should be understood that a limit is something different from the value of a function and it is only an afterthought if we wish to define the value of function using its limit.

Another point you should understand is that the derivative $f'(x)$ is not the value of $g(h)$ at $h=0$, but it is defined to be the limit of $g(h)$ as $h\to 0$. The derivative does not exist separately as the slope of tangent at $x$, rather the slope of tangent is defined to be the derivative $f'(x)$ which is further defined as a limit.

• Thanks for the answer. I intuitively think about the slope between the points $x$ and $x+h$ as defined as $\dfrac{f(x+h)-f(x)}{h}$ and I tend to think about the slope of the tangent on the point $x$ as the result of the function $g(0)$ if we would able to evaluate it. But as you have said, this exact location is troublesome and assuming it exists, we follow the limit of the function around the point $h=0$ in order to infer what would be the value of the slope at $h=0$ if it were defined in the domain of $g(h)$. – Ufuk Can Bicici Sep 4 '14 at 14:23
• (About your last edit): Derivative is defined as a limit but as the physical meaning it gives us the slope of the tangent line at $x$, isn't that right? As in my first comment, we can always calculate an erroneous, approximate tangent slope by $g(h)=\dfrac{f(x+h)-f(x)}{h}$ and if some magic would happen and we could define it at $h=0$, it would give us the value of the slope, hence the value of the derivative. Since we can never define this function that way, we analyze its behavior around the close proximity of $h=0$ and deduce that it would gave the limit value $L=f'(x)$ at $h=0$ ... – Ufuk Can Bicici Sep 4 '14 at 14:35
• ... if it were defined. The limit definition arises because of our inability to evaluate a division by zero. Isn't this thought right? – Ufuk Can Bicici Sep 4 '14 at 14:36
• @UfukCanBiçici: Slope of curve is always defined as a limit. Slope of a straight line can be defined as $(b-d)/(a-c)$ if line has two distinct points $(a,b)$ and $(c,d)$. – Paramanand Singh Sep 4 '14 at 14:40
• @UfukCanBiçici: you need to get used to this approach and then u will find it less strange! Was really nice to have this discussion with you. Thanks! – Paramanand Singh Sep 4 '14 at 14:55