Why do I rationalize the numerator in this question? I've been going through the diagnostic tests for my Calculus Textbook to get ready for classes starting on Monday. One of the questions is this:

Rationalize the expression and simplify.
$\frac{\sqrt{4+h}-2}{h}$

The answer is $\frac{1}{\sqrt{4+h}+2}$.
I understand how to get to that equation, but I don't understand why. Wasn't the denominator of the equation already rational? What's the point of rationalizing the numerator in this case?
 A: I don't like the term rationalize here. What you've really done is put the expression in a form where plugging in zero for $h$ makes sense: you can't do this in the first form because you can't divide by zero. That is likely what the textbook author's intent was. I think they created this as a silly pre-exercise to what you'll be doing later so you'll "be prepared" to evaluate limits. Soon you'll do something that looks like
$$
  \lim_{h \to 0}\frac{\sqrt{4+h}-2}{h} = 
  \lim_{h \to 0}\frac{1}{\sqrt{4+h}+2} = 
  \frac{1}{\sqrt{4+0}+2} = \frac{1}{4}
$$
A: I agree that asking you to "rationalise" is unhelpful.
But there is a very good reason why the second fraction is preferable to the first.
In calculus we often have to evaluate the behaviour of expressions as some variable (often $h$) gets very close to zero.
The first expression $\frac{\sqrt{4+h}-2}{h}$ is approximately $\frac{\sqrt{4+0}-2}{0} \approx \frac 00$ which is not well-defined.
The second expression $\frac{1}{\sqrt{4+h}+2}$ is approximately $\frac{1}{\sqrt{4+0}+2} \approx \frac 14$ - much nicer!
A: We rationalize  numerator (vs. denominator) since it removes an apparent singularity at $\,h=0$.
For example, one can make the quadratic formula work even in the degenerate case when the lead coefficient $\,a = 0\,$ by rationalizing the numerator as below 
$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\ =\  \dfrac{2c}{-b \pm \sqrt{b^2-4\:a\:c}}$$
As $\,a\to 0,\,$ the latter yields the root $\,x = -b/c\ $ of $\ bx+c\,\ (= ax^2\!+bx+c\,$ when $\,a=0).$
Below is a generalization of your problem that I posted in a closely related question.
If $\rm\ f(x)\: = \ f_0 + f_1\ x +\:\cdots\:+f_n\ x^n\:$ and $\rm\: f_0 \ne 0\:$ then rationalizing the numerator below yields
$$\rm \lim_{x\:\to\: 0}\ \dfrac{\sqrt{f(x)}-\sqrt{f_0}}{x}\ = \ \lim_{x\:\to\: 0}\ \dfrac{f(x)-f_0}{x\ (\sqrt{f(x)}+\sqrt{f_0})}\ =\ \dfrac{f_1}{2\ \sqrt{f_0}}$$
Your problem is the special case  $\rm\ f(x) = 4 + x\ $ with $\rm\ f_0 =4,\ f_1 = 1\:,\:$ so the limit equals $\:1/4\:.\:$  
When you study derivatives you'll see how they mechanize this process in a very general way. Namely the above limit is $\rm\:g'(0)\ $ for $\rm\:g(x) = \sqrt{f(x)}\:,\:$ so applying general rules for calculating derivatives we easily mechanically calculate that $\rm\:g'(x)\: =\: f\:\:'(x)/(2\:\sqrt{f(x)})\:.\:$ Evaluating it at $\rm\:x=0\:$ we conclude that $\rm\: g'(0)\: =\: f\:\:'(0)/(2\:\sqrt{f(0)})\: =\: f_1/(2\:\sqrt{f_0})\:,\:$ exactly as above.
A: $$\frac{\sqrt{4+h}-2}{h}=\frac{\sqrt{4+h}-2}{h}\times \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}=\frac{4+h-4}{h(\sqrt{4+h}+2)}=\frac1{\sqrt{4+h}+2}$$
