Is the limits technique just an approximation to find derivatives of functions? If $\displaystyle \lim_{x \to \Delta}$ means that as x approach $\Delta$ but never reach it, that is $\Delta - x \neq  0$, so the techniques to find the derivative of a function is just an approximation and not really the derivative since if we get when calculating a derivative of a function: $\displaystyle \lim_{x \to \Delta} x^2 + x \Delta$ then if $\Delta \neq 0$ then it is just an approximation since a little number will remain, where I'm wrong?
I think something that gets ignored when talking about limits is just how much we use the Squeeze Theorem:  This says that if $f(x)\leq g(x)\leq h(x)$ for all $x$ in an open interval around $x=a$ and
$$
\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L
$$
then $\lim_{x\to a}g(x)=L$.  As in @Keeran's answer, we can let $f(x)=x^2$.  We obtain
\begin{align*}
f'(x) &amp;= \lim_{h\to 0}\frac{(x+h)^2-x^2}{h}\\
     &amp;=  \lim_{h\to 0}\frac{2xh+h^2}{h}\\
     &amp;=  \lim_{h\to 0}(2x+h)
\end{align*}
Before we plug in $h=0$ there is a point to be made.  In that last step we are not saying that, as functions of $h$, 
$$
\frac{2xh+h^2}{h}=2x+h.
$$
We are saying that the limits
$$
\lim_{h\to 0}\frac{2xh+h^2}{h}
$$
and
$$
\lim_{h\to 0}(2x+h)
$$
are equal.  This is the Squeeze Theorem at work.  

In your question you ask whether we have an approximation.  In answer to that: Up until we let $\Delta=0$, yes, we only have an approximation.  My point here is that the derivative is defined to be the limit above if it exists.  The Squeeze Theorem is what allows us to figure out the limit.
In taking the limit the variable that you set the limit to does reach the given value. If you are thinking of the definition of the derivative as follows (for functions on the real numbers)

$\displaystyle f'(x) = \lim_{h\rightarrow 0} {f(x+h) - f(x) \over h }$

then we do take $h=0$ at  some point. For example consider taking the derivative of the function

$\displaystyle f(x) = x^{2}$

using the above definition. We have

$\displaystyle f'(x) = \lim_{h\rightarrow 0} {f(x+h) - f(x) \over h} \\
\displaystyle \quad\quad\ = \lim_{h\rightarrow 0} {x^{2} + 2hx + h^{2} - x^{2} \over h} \\
\displaystyle \quad\quad\ =\lim_{h\rightarrow 0} 2x + h$

Now set $h=0$ and you have your derivative, i.e. $\displaystyle{\text{d}\over\text{d}x}x^{2} = 2x$