Well, in a sense, you're right. When they say that the limit of $f(x)$ at $x=a$ is $L$ it doesn't necessarily mean that $f(a)=L$. Actually, the nice idea behind limits is that you can talk about the limit of a function even if the function is not defined at that value. This is a very powerful idea that later enables us to talk about derivatives as you possibly know.
For example $\displaystyle \lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$ but the value of $\displaystyle \frac{\sin(x)}{x}$ is not defined at $x=0$. If you graph it on wolframalpha, you'll see that this means 'as we approach $x=0$ the value of $\displaystyle \frac{\sin(x)}{x}$ approaches 1'. We never claim that these two are equal! We just claim that the value of $f(x) = \displaystyle \frac{\sin(x)}{x}$ can become arbitrarily close to $1$ provided that we let $x$ be close enough to $0$ .
When we say that the limit of $f(x)$ at $x=a$ is $L$, we are claiming that we can make $f(x)$ arbitrarily close to $L$ provided that we take $x$ close enough to $a$. That's all.
So, let's review the definition of a secant line, the definition of a tangent line, and the definition of the derivative:
A secant line to a curve is a line that intersects the curve in two distinct points.
The derivative of a function $f(x)$ at point $a$, denoted $f'(a),$ is defined to be the limit
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}, $$
provided this limit exists.
A tangent line to a curve $f(x)$ at a point $a$, if it exists, is defined to be that line with slope $f'(a)$ going through the point $(a, f(a)).$
(Side note: a tangent line is most definitely NOT a line that intersects a curve at only one place. This is a common, but misguided definition of the tangent line often encountered in high school geometry textbooks and is incorrect in general, though true for conic sections such as circles, ellipses, hyperbolas, and parabolas. It's better to think of tangent lines in terms of the derivative.)
First off, notice that the derivative is defined as a limit. So, for the derivative to exist, the limit has to exist. There are multiple ways a limit in general can fail to exist (right-hand and left-hand limits not equal, infinities, etc.), and there are additional ways that a derivative-type limit can fail to exist. For example, while this is not sufficient for the derivative to exist, the function must be continuous at $a$. That means $\lim_{x\to a}f(x)=f(a),$ a statement which says three non-trivial things: 1. The limit exists. 2. $f(a)$ is defined. 3. The limit equals the value of the function. Any of those can fail. In addition, you could have, say, a corner at $a$. The function $f(x)=|x|$ has a corner at $a=0$. While the function is continuous there,
$$1=\lim_{x\to 0^{+}}\frac{|x|-|0|}{x-0}\not=\lim_{x\to 0^{-}}\frac{|x|-|0|}{x-0}=-1.$$
The function $f(x)=|x|$ is therefore not differentiable at $x=0,$ because the required limit does not exist. Geometrically, think about the corner this way: what unique value could you assign to the "slope" of $|x|$ at $x=0?$ You could draw any number of candidates. What the limit definition is saying is that you have to be able to come up with a unique value for the slope at a point, in order for the function to be differentiable at that point.
As for the relationship between secant lines and tangent lines, just notice that the slope of the secant line to $f$ at the points $x$ and $a$ is given by
$$m=\frac{f(x)-f(a)}{x-a}.$$
The slope of the tangent line to $f(x)$ at $x=a$ is defined as the limit of the secant line slopes. You can imagine, geometrically, that you're letting the point $x$ get arbitrarily close to $a$. In the limit, you get the tangent line (provided the limit exists).
Best Answer
You're right. We don't ever reach that point. We take a limit.
The colloquialism, "reaching the point" is a good anthropomorphic description. Limits allow us to stretch the constraints of the real numbers by pushing towards the infinite and infinitesimal. Technically, though, to venture into such territory, we need to properly define limits. This is often introduced with the epsilon-delta formalization.
Say there exists a limit $f'(x)=\lim_{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$. Then for every $\epsilon>0$, there exists some $\delta>0$ such that whenever $0<\Delta x<\delta$, we find $|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$.
We can heuristically think of the last paragraph as the following: our derivative exists if for every positive number $\epsilon$ and $\delta$, including the most ridiculously small numbers you can ever imagine, whenever $\Delta x$ is trapped between zero and any of these ridiculously small numbers, the difference between our derivative and the original expression is imperceptible.
But wait a minute, you say
The epsilon-delta definition seems to hint that as well, but there's a catch: $$|f'(x) - \frac{f(x+\Delta x)-f(x)}{\Delta x}|<\epsilon$$
This is not less than some real positive number $\epsilon$. This is less than ANY POSSIBLE real positive number $\epsilon$. Such a concept only exists within the formalism of a limit, and is by no means a measurable quantity. That's what is meant by infinitesimal.
Due to the limit, then, the derivative cannot represent any possible secant line. There are no two points corresponding to $x+\Delta x$ and $x$ that are indistinguishable! The value we reach has converged to that which represents the slope of the tangent.
Added note: $\Delta x\rightarrow 0$ doesn't just imply that $\Delta x$ is running through the positive numbers towards zero. For the limit to exist, we typically require it to be two-sided, meaning that $\Delta x\rightarrow0^+$ and $\Delta x\rightarrow0^-$ must produce the same result. In either case, the difference between $\Delta x$ and zero becomes vanishingly small.