In this series, I'm trying to convey an intuitive understanding of differential calculus, so I'm deliberately not completely mathematically accurate. Still, it was time for a little more pedantry at this point, especially with regard to the notions of limit transition and derivative presented in the last chapter. So let's talk about them in more detail and back them up with more solid reasoning and logic so that we can really use them in square root estimates. Surely knowledge of these concepts will prove valuable in the way ahead.

One limit transition process may look like this in mathematical notation, for example: $$ \lim_{h\to 0} 1 + h = 1 \,.$$

We're saying that the limit for $h$ going to zero of $1+h$ is equal to one. The concept of limit established here is that we are making a limit transition. The fact that $h$ goes to zero means that for an arbitrary small number, $\epsilon$, the $h$ is less than the $\epsilon$. The following dialogue could be made:

- Let's have $h$ going to zero.
- Does that mean $h$ is zero?
- Well, $h$ is not zero. But with any great degree of accuracy, we can say it's zero.
- What?
- Well, for any need anyone can ever invent, $h$ can't be distinguished from zero. Let's say $h$ is some distance in centimeters. Then no one can tell the difference between $h\,\text{cm}$ and $0\,\text{cm}$.
- I mean, I can take a ruler and measure the distance $h$.
- Yes, but there are pieces on the ruler. The smallest piece is one millimeter. Our $h$ happens to be less than half a millimeter, so it's closer to zero than the smallest piece. So we measure zero centimeters.
- Well, I'll take the micrometer...
- Well, on a micrometer, the smallest piece is one micrometer. Our $h$ happens to be less than half a micrometer in micrometers, so we receive zero again. And no matter what other instrument you take to measure distance, $h$ is so small that it would always show that $h$ equals zero.
- So what's the difference between $h$ and zero?
- Well, $h$ centimeters is still some distance between two different points, so it's non-zero. It's just a very small distance. It may seem a little strange at first, but you'll soon see how much use it has.

Let's quickly calculate a few limits for greater understanding. First example: $$ \lim_{h\to 0} (x-h)^2 = x^2 \,.$$

Here $h$ goes to zero. The first thing to try when evaluating a limit is to put down $h$ equal to zero. It may be that we get a straight result like this. Second example: $$ \lim_{h\to 0} \frac{h^2}{h} =\,?$$

We can try to substitute $h=0$ first. But then we divide by zero, and you can't do that! Fortunately, we can simplify the fraction and get: $$ \lim_{h\to 0} \frac{h^2}{h} = \lim_{h\to 0} h = 0 \,.$$

The advantage of limit transitions has now been demonstrated: they allow us to work with terms that would not normally be legal\u2019. In short, $h$ has all the advantages of zero (the expression is simpler) and no drawbacks (that you can't divide by zero). Third example: $$ \lim_{h\to 0} \frac{1}{|h|} = \infty \,.$$

Here we see that $|h|$ is some very small positive number. The smaller it is, the larger the resulting limit. So as $|h|$ goes to zero, so the term in limit grows beyond all limits and that's why we write that it equals infinity.\

But there is not always a limit. For example, $\lim_{h\to 0} 1/h$ does not exist because we are unable to decide if it should be $+\infty$ or $-\infty$. Depends on which side $h$ would be close to zero.

That's all in terms of limit examples. It was a short crash course in which it was simply a matter of understanding the principle of it. More on limits only when we really need them.

A good way to remember the definition of a derivative is by the simplified form $\frac{\Delta y}{\Delta x}$, or changing $y$ to change $x$.

The derivative by which we approximated the function at some points actually has the following definition (the equals sign with three commas means we define): $$\frac{\mathrm{d} f}{\mathrm{d} x} \equiv \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \,.$$

Isaac Newton developed his theory of differential calculus in England in the 17th century. At a similar time, the German Leibniz published his theory of differential calculus, which dealt with the same, only using different markings. Newton kept his theory in a drawer, but after seeing that Leibniz had published on a similar subject, he rushed to publish it too. There was a fierce debate about who actually invented calculus first, which in some circles continues to this day.

Leibniz's marking makes sense in that $\mathrm{d} x$ equals $\Delta x$ and $\mathrm{d} f(x)$ equals $\Delta y$. If $\Delta x$ was a small offset of $x$, then according to Leibniz, $\mathrm{d} x$ is a very small offset of $x$. Even infinitesimally small.

Other than the limits before the fraction, this definition is practically the same as last chapter, only we have $h$ instead of delta. A good way to remember a derivative is to remember that the derivative is roughly $\Delta y/\Delta x$, change in functional value to change $x$. See Figure.

We're finally ready for the square root. Let's try to work out what roughly $\sqrt{65}$ is. Let's start from point $x_0 again = $64 for function $f(x)=\sqrt{x}$. So we need to figure out the derivative of this function, so let's use the definition: $$\frac{\mathrm{d} f}{\mathrm{d} x}= \lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \,.$$

And we're immediately running into a problem, because if we just put $h=0$, we're dividing by zero, and you can't do that. So we have to figure out how to adjust the fraction so that there is no longer $h$ in the denominator. Think about what might help for a while and maybe you can figure it out with a pencil and paper.

Hint: we want to get rid of $h$ under the square root.

Help #2: let's use the formula $(a+b)(a-b)=a^2-b^2$.

Solution: $$\begin{align*} \lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} &= \lim_{h\to 0} \frac{\left(\sqrt{x+h} - \sqrt{x}\right) \left( \sqrt{x+h} + \sqrt{x}\right)}{h\left( \sqrt{x+h} + \sqrt{x}\right)} \\ &= \lim_{h\to 0} \frac{h}{h\left( \sqrt{x+h} + \sqrt{x}\right)} \\ \frac{\mathrm{d} f}{\mathrm{d} x} &= \frac{1}{2\sqrt{x}} \,. \end{align*}$$

$$f(65) \approx f(64) + f'(64)\cdot 1 = 8 + 1/(2\cdot 8) = 8+1/16 = 8,0625 \,.$$

The real value at this is about $8,0623.

Before we get to the end of this chapter, let's look at the derivative of the important function class $f(x)=x^n$ for some $n$. So let's get into the definition of a derivative: $$\frac{\mathrm{d} f}{\mathrm{d} x}= \lim_{h\to 0} \frac{(x+h)^n - x^n}{h} \,.$$

We obviously need to modify the expression $(x+h)^n$. Fortunately, we have a chapter on the Pascal Triangle from which to write: $$(x+h)^n = x^n + n\cdot h\cdot x^{n-1} + h^2 \cdot(\dots ) \,.$$

Here I have written the dots instead of some final positive expression that may no longer concern us. Why? We'll know when we continue to define a derivative: $$\begin{align*} \frac{\mathrm{d} f}{\mathrm{d} x}&= \lim_{h\to 0} \frac{x^n + n\cdot h\cdot x^{n-1} + h^2 \cdot(\dots ) - x^n}{h} \\ \frac{\mathrm{d} f}{\mathrm{d} x}&= \lim_{h\to 0} n\cdot x^{n-1} + h \cdot(\dots ) \\ \frac{\mathrm{d} f}{\mathrm{d} x}&= nx^{n-1} \,. \end{align*}$$

This formula applies to all full $n$, however it also applies to all other $n$ (real, complex...). For more general cases, we are proving claims similarly, however I will not be taking evidence as it would take somewhat more time.