Comparison of increments

Consider the function

$$y = x^2.$$

Assuming a fixed initial value for $x$, let $x$ take on an increment $\Delta x$. Then $y$ will take on a corresponding increment $\Delta y$, and we have

$$y + \Delta y = (x + \Delta x)^2,$$

or,

$$y + \Delta y = x^2+ 2x \cdot \Delta x + (\Delta x)^2.$$

Subtracting $y=x^2$ from this,

$$\Delta y = 2x \cdot \Delta x + (\Delta x)^2,$$ (4.1)

we get the increment $\Delta y$ in terms of $x$ and $\Delta x$. To find the ratio of the increments, divide (4.1) by $\Delta x$, giving

$$\frac{\Delta y}{\Delta x} = 2x + \Delta x.$$

If the initial value of $x$ is $4$, it is evident that

$$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = 8.$$

Let us carefully note the behavior of the ratio of the increments of $x$ and $y$ as the increment of $x$ diminishes.

Initial	New	Increment	Initial	New	Increment
value of $x$	value of $x$	$\Delta\, x$	value of $y$	value of $y$	$\Delta\, y$	$\frac{\Delta y}{\Delta x}$
4	5.0	1.0	16	25.	9.	9.
4	4.8	0.8	16	23.04	7.04	8.8
4	4.6	0.6	16	21.16	5.16	8.6
4	4.4	0.4	16	19.36	3.36	8.4
4	4.2	0.2	16	17.64	1.64	8.2
4	4.1	0.1	16	16.81	0.81	8.1
4	4.01	0.01	16	16.0801	0.0801	8.01

It is apparent that as $\Delta\, x$ decreases, $\Delta\, y$ also diminishes, but their ratio takes on the successive values $9$, $8.8$, $8.6$, $8.4$, $8.2$, $8.1$, $8.01$; illustrating the fact that $\frac{\Delta y}{\Delta x}$ can be brought as near to $8$ in value as we please by making $\Delta\, x$ small enough. Therefore^4.3,

$$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = 8.$$