$\varepsilon-\delta$ definition - Is the graph accurate? 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? 
 A: 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.

In other terms,
$$f([x_0-\delta,x_0+\delta])\subseteq[L-\epsilon,L+\epsilon],$$ not
$$f([x_0-\delta,x_0+\delta])= [L-\epsilon,L+\epsilon].$$
A: 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).
