TITLE: Curiosity: Wouldn't the definition of the derivative always be 1 if it exists?
QUESTION [10 upvotes]: I'm pretty sure I have the wrong intuition here but I have a slight confusion about the way we could calculate the derivative at a certain point using (one of) the definition(s) of the derivative. See example bellow:
$$\frac{df(x)}{dx}= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
let's see the case of $f(x) = \sqrt{5x+10}$
$$\frac{df(x)}{dx}=\lim_{h\to0}\frac{\sqrt{5h+5x+10}-\sqrt{5x+10}}{h}$$
If we want to calculate $f'(5)$
$$\left.\frac{df(x)}{dx}\right\rvert_5=\lim_{h\to0}\frac{\sqrt{5h+35}-\sqrt{35}}{h}$$
if we try to find the limits when $h\to0^+$:

The numerator would be only slightly superior to 0
The denominator would be only slightly superior to 0

$$\frac{\text{very small number above zero}}
       {\text{very small number above zero}}\approx 1$$
It should be the same for $h\to 0^-$
Hence: 
$f'(5)= 1$?
N.B: I know this result is wrong, I just want to know how the logic I used is faulty.

REPLY [4 votes]: $\lim_{h\to0}$ doesn't really make any statement like “$h$ is a particular very small number”. Rather it considers its argument as a function of a general $h$, decides how small $h$ must be so that function becomes practically constant, and then yields that constant. In principle that doesn't require evaluating for any very small values at all, for example with
$$
  f(x) = \begin{cases}1&\text{for }x<1\\x & \text{else}\end{cases}
$$
we have
$$
  \lim_{x\to0}f(h) = f(0.0001) = f(0.5) = f(0) = 1
$$
because this function is really constant if only $x$ is smaller than 1. In practice, limits aren't usually over such functions that are constant in a whole region, but they are necessarily over continuous functions, which guarantee you that you can make the error-from-constness arbitrarily small by going to sufficiently small values; but the arguments are still always proper real numbers – no “this doesn't behave like numbers otherwise do”!
And for an ordinary “somewhat small” $h$, say $h=0.5$, you'd certainly agree that $\frac{\sqrt{5\cdot h + 35} - \sqrt{35}}{h}\neq 1$. In fact any pocket calculator will tell you that it is $\approx0.415$. If you then make $h$ yet smaller, the following will happen (on a computer with IEEE754 double precision arithmetic):
$$\begin{align}
\\ 10^{-1} &&& 0.4210786080687612
\\ 10^{-2} &&& 0.4224263146657137
\\ 10^{-3} &&& 0.4225620364017857
\\ 10^{-4} &&& 0.422575618168608
\\ 10^{-5} &&& 0.42257697643321984
\\ 10^{-6} &&& 0.42257711196924674
\\ 10^{-7} &&& 0.4225771199628525
\\ 10^{-8} &&& 0.42257708443571573
\\ 10^{-9} &&& 0.4225766403465059
\\ 10^{-10} &&& 0.4225775285249256
\\ 10^{-11} &&& 0.4225952920933196
\\ 10^{-12} &&& 0.4227729277772596
\\ 10^{-13} &&& 0.41744385725905886
\\ 10^{-14} &&& 0.4440892098500626
\\ 10^{-15} &&& 0.8881784197001251
\\ 10^{-16} &&& 0.0
\\ 10^{-17} &&& 0.0
\\ 10^{-18} &&& 0.0
\\ 10^{-19} &&& 0.0
\\ 10^{-20} &&& 0.0
\end{align}$$
Notice how the “moderately small” arguments give a very consistent value of 0.4225something – this corresponds to the actual “exact derivative”. But extremely small arguments suddenly give complete bogus. This is similar to your question: with extremely small numbers, the computer can't really calculate anymore (it basically runs out of digits to store the deviations in), so you then have a $0 \stackrel{?}= \frac00 \stackrel{?}= 1$ kind of situation.
Well – one could say that this is only an artifact of the floating-point processor. But IMO it gets to the very heart of how analysis works: it exploits the fact that certain functions behave in a certain regime (often close to a singularity of utter indeterminacy!) very predictable, so they can be well approximated by something simpler. That can then be used for further calculations, which would otherwise have been unfeasible. While it can be mathematically useful to describe the deviations as infinitesimally small, especially for physical applications it's actually more appropriate to say just small enough so we don't have to worry about higher-order effects, but not so small that the signal is lost in the noise floor.