The confusion seems to be in your line
$\lim\limits_{x\to a}{f(x)}$ approaches $L$.
It is $f(x)$ that approaches $L$.
That is, the function $f(x)$ approaches some value as $x$ tends to $a$. That value is called the limit, and is denoted $\lim\limits_{x\to a}{f(x)}$. Saying $\lim\limits_{x\to a}{f(x)}=L$ is saying these two values are equal.
The derivative $\frac{dy}{dx}$ is the gradient of the tangent line, which is the limit of gradients of successively smaller secants. Again, a limit is a value, not something that approaches a value.
Edit: let me see whether I can help with your rewritten question.
Suppose $f$ is a continuous function. Then $\lim\limits_{x\to a}f(x) = f(a)$ (this is by definition of continuity -- we haven't taken derivatives yet).
Now suppose $f$ is differentiable at $a$, and suppose $f'(a) = L$ -- that is, $$\lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h} = L$$.
The function $$g(h) = \frac{f(a+h)-f(a)}{h}$$, which is defined for $h$ non-zero, represents the gradients of secants. The function $g$ approaches $L$ as $h$ tends to $0$ -- this is what it means for $f$ to be differentiable at $a$. So $g$ approaches the gradient of the tangent to the curve at $a$. But $g$ is not the derivative. The derivative is $\lim\limits_{h\to 0}g(h)$, which is $L$.
The symmetric derivative being defined at more places isn't a good thing.
In my mind, the main point of differentiation is to locally approximate a function by a linear function. That is, the heart of saying that the derivative $f'(a)$ exists at a point $a$ is the statement that
$$f(x) = f(a) + f'(a) (x - a) + o(|x - a|)$$
as $x \to a$, and if I were the King of Calculus this is how the derivative would actually be defined. (Among other things, this definition generalizes smoothly to higher dimensions.) Removable discontinuities are a non-issue as they should just be removed, but at a cusp we do not have this property for any possible value of $f'(a)$, so we shouldn't be talking about derivatives at such points at all. (We can talk about left or right derivatives, but this is something different.)
The symmetric derivative at $a$ is not a natural definition. It has the utterly strange property that any weirdness in a neighborhood of $a$ is ignored if it happens to be canceled by equivalent weirdness after reflecting around $a$. Let me give an example. Consider the function $f(x) = 1_{\mathbb{Q}}(x)$ which is equal to $1$ if $x$ is rational and $0$ otherwise. The symmetric derivative of $f$ at any rational point exists and is equal to $0$! Is there any reasonable sense in which $f$ is differentiable at a rational point?
The ordinary derivative, on the other hand, is sensitive to weirdness around $a$ because it compares all of that weirdness to $f(a)$.
Best Answer
As $x+h\to x\implies (x+h)-x\to0\implies h\to0$. Does that make you understand now?