Part 3/5:
Here, the quantity ( \epsilon ) becomes vital as it is defined in such a way that it vanishes when ( \Delta x ) approaches zero. Thus, as ( \Delta x ) nears zero:
[ \lim_{\Delta x \to 0} \epsilon = 0 ]
The Core of the Proof
From the earlier definitions, we rearrange ( \Delta y ) to write it in terms of ( \epsilon ):
[ \Delta y = (f'(x) + \epsilon) \Delta x ]
Now, introducing the function in terms of another variable, we state:
[ y = f(u) \quad \text{and} \quad u = g(x) ]
The increment for ( \Delta y ) can therefore be defined as:
[ \Delta y \sim \Delta u \cdot f'(u) + \epsilon_1 ]
where ( \Delta u ) appropriately reflects the change in terms of ( x ).