Reason for relaxing limit in derivative in this context?

Question

Consider the following paragraph from NUMERICAL COMPUTATION of the deep learning book..

Suppose we have a function $y = f(x)$, where both $x$ and $y$ are real numbers. The derivative of this function is denoted as $f'(x)$ or as $\dfrac{dy}{dx}$. The derivative $f'(x)$ gives the slope of $f(x)$ at the point $x$. In other words, it specifies how to scale a small change in the input to obtain the corresponding change in the output: $f(x+\epsilon) \approx f(x) + \epsilon f'(x)$.

I have doubt in the equation $f(x+\epsilon) \approx f(x) + \epsilon f'(x)$ given in the paragraph.

In strict sense, the derivative function $f'$ of a real valued function $f$ is defined as

$$f'(x) = \lim_{\epsilon \rightarrow 0} \dfrac{f(x+\epsilon)-f(x)}{\epsilon}$$

wherever the limit exists.

If I replace the original definition of the derivative as follows

$$f'(x) \approx \dfrac{f(x+\epsilon)-f(x)}{\epsilon}$$

then I can obtain the equation given in the paragraph i.e, $f(x+\epsilon) \approx f(x) + \epsilon f'(x)$.

But, my doubt is that how can I modify the definition with $\lim\limits_{\epsilon \rightarrow 0}$ to an approximation with out limit? How can the following two are same?

$$f'(x) = \lim_{\epsilon \rightarrow 0} \dfrac{f(x+\epsilon)-f(x)}{\epsilon} \text { and } f'(x) \approx \dfrac{f(x+\epsilon)-f(x)}{\epsilon}$$

Answer 1

The equation $$f(x + \epsilon) \approx f(x) + \epsilon f'(x)$$ is justified from taylor series. It is not derived from the limit definition of a derivative. Let $f'(a)$ and $f''(a)$ exist for $a$ in the interval $(x, x+\epsilon)$. Then, \begin{align*} f(x + \epsilon) = f(x) + \epsilon f'(x) + \frac{\epsilon^2}{2}f''(b) \end{align*} where $b$ is some number between $(x, x + \epsilon)$. If $\epsilon << 1$, then $\frac{\epsilon^2}{2}f''(b)$ is small so it can be ignored, leading to the approximation.

taylor's theorem

Answer 2

It is just an assumption. If we assume $\epsilon$ is small enough (depending on the function $f$), you can remove the limit $\lim_{\epsilon \to 0}$ for the approximation.