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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?
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 $$ \lim_{h \to 0}\frac{\sqrt{4+h}-2}{h} = \lim_{h \to 0}\frac{1}{\sqrt{4+h}+2} = \frac{1}{\sqrt{4+0}+2} = \frac{1}{4} $$
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!
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.
$$\frac{\sqrt{4+h}-2}{h}=\frac{\sqrt{4+h}-2}{h}\times \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}=\frac{4+h-4}{h(\sqrt{4+h}+2)}=\frac1{\sqrt{4+h}+2}$$
