# Question about infinitesimals and approximations

#### biometrix

##### New member
Hello.

I am reading about infinitesimals and came up to the following calculation: (Suppose f is differentiable at point x = a)

$$\displaystyle Δf - dy = Δf -f'(a)Δx = (\frac{Δf}{Δx} - f'(a))\cdot Δx = ε \cdot Δx$$

Now as Δx approaches 0 the fraction $$\displaystyle \frac{Δf}{Δx}$$ approaches f'(a) and therefore the portion inside the parentheses approaches 0. (Based on the textbook)

But the definition of the fraction $$\displaystyle \frac{Δf}{Δx}$$ as Δx approaches 0 is exactly f'(a).

So why is it that ε approaches 0 instead of ε equals to 0.

Thanks.

#### pka

##### Elite Member
I am reading about infinitesimals and came up to the following calculation: (Suppose f is differentiable at point x = a)
$$\displaystyle Δf - dy = Δf -f'(a)Δx = (\frac{Δf}{Δx} - f'(a))\cdot Δx = ε \cdot Δx$$
Now as Δx approaches 0 the fraction $$\displaystyle \frac{Δf}{Δx}$$ approaches f'(a) and therefore the portion inside the parentheses approaches 0. (Based on the textbook) What is the textbook?
But the definition of the fraction $$\displaystyle \frac{Δf}{Δx}$$ as Δx approaches 0 is exactly f'(a).
So why is it that ε approaches 0 instead of ε equals to 0.
Without knowing the exact textbook ti is impossible to answer.
Here is a free text book that was one used for infinitesimal calculus,
In Elementary Calculus: An Infinitesimal Approach
In Elementary Calculus: An Infinitesimal Approach by Jerome Keisler in chapter 3, there is good proof of this. The chapters and whole book is a free down-load at http://www.math.wisc.edu/~keisler/.