# Reason for relaxing limit in derivative in this context?

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?

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

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.
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.