$\varepsilon-\delta$ definition - Is the graph accurate?

Question

I am going through an introductory course on Calculus.

In chapter 2, section 2.4, p 111, the author presents the following graphs to illustrate the $\varepsilon-\delta$ limit.

I have understood the epsilon-delta definition but I think that the fig 5 and 6 are inaccurate.

Bare with me as I am trying to understand the material better.

The correct fig should look like this (notice the red horizontal line, which indicates $y=L+\varepsilon$):

Edit::

The rationale behind this is as follows:

The function $f$ maps all the points in the interval $(a-\delta, a+\delta)$ into the interval $(L-\varepsilon, L+\varepsilon)$. So $f(a-\delta)$ should lie below the $y=L+\varepsilon$.

Am I right?

Answer 1

The legend means: "when $x$ is in here, $f(x)$ is in here" and the figure is right.

The converse condition "when $f(x)$ is in here, $x$ is in here" is not required.

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In other terms,

$$f([x_0-\delta,x_0+\delta])\subseteq[L-\epsilon,L+\epsilon],$$ not $$f([x_0-\delta,x_0+\delta])= [L-\epsilon,L+\epsilon].$$

Answer 2

Let's understand this from definition of limits itself. We say that $\lim_{x\to a}f(x) =L$ if
For every $\epsilon \gt 0$ there exists a $\delta\gt0$ such that if $x$ satisfies $0<|x-a|<\delta$, then $|f(x) - L|\lt \epsilon $
Now note that the figures 5 and 6, which you may think to be inaccurate are not inaccurate because (see the figures 5 and 6)if $x$ satisfies $0<|x-a|\lt \delta$ then $|f(x)-L| \lt \epsilon$
On the other hand, you are considering the converse (if f satisfies $|f(x) - L|<\epsilon$ then $0<|x-a|<\delta$) of the above implication. And converse of an implication is not necessarily true.
To be more specific, if you consider the red horizontal line, you are considering an another $\epsilon \gt 0$ (lesser than the previous epsilon, as per fig 5),for which $\delta>0 $ will be different (lesser than previous delta, as per fig 5).