Finding $\frac{\mathrm d}{\mathrm dx}(6)$ by first principles I had an exercise in my text book:

Find $\frac{\mathrm d}{\mathrm dx}(6)$ by first principles.

The answer that they gave was as follows:
$$\lim_{h\to 0} \frac{6-6}{h} = 0$$
However surely that answer would be undefined if we let $h$ tend towards $0$, as for finding all other limits like that we substitute $0$ in the place of $h$? 
 A: The answer is right. We know that if the following limit $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ exists, then it is denoted by $f'(x)$. In the case $f$ is a constant function, the numerator of previous fraction in limit is absolutely zero but the variable $h$ in the denominator is going to zero but it is not zero. We know many functions that $x=0$ is excluded from their domains but we prove their limit at zero.
A: When evaluating the limit:
$$\lim_{h\to 0} \frac{6-6}{h}$$
notice that (provided that $h \ne 0$), we have:
$$\frac{6-6}{h}=\frac{0}{h}=0$$
Thus, since it is possible to evaluate $\lim_{x\to a}f(x)$ even if $f(a)$ is undefined, we have:
$$\lim_{h\to 0} \frac{6-6}{h} = \lim_{h\to 0} 0 = 0$$
A: In all cases in which the $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ exists, both the numerator and the denominator approach $0$.  And even if the limit does not exist the numerator and denominator both become $0$ when $h=0$.
But $\lim\limits_{h\to0}$ of anything depends only on the behavior of that function of $h$ when $h$ is not $0$.
These are the most important things you can learn about limits before getting into these problems about derivatives.  If these things were not as they are, the reason for learning about limits before going on to derivatives just wouldn't be there at all.
