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.


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?

  • 1
    $\begingroup$ I believe it was a simple exercise to test your ability in rationalising expressions. There is no real other explanation, to me. $\endgroup$
    – Enrico M.
    Commented Aug 27, 2016 at 21:46
  • $\begingroup$ The aim is to calculate the limit when $h\to2$. As it is given, this is an indeterminate form. $\endgroup$
    – Bernard
    Commented Aug 27, 2016 at 21:53
  • $\begingroup$ I agree with your irritation. When one hears "rationalize" it usually means to rationalize the denominator (although why in the heck that's important is something I've never understood). So to say "rationalize" seems ... ambiguous. But the second part is "simplify" which... means to reduce the number of occurences of the varible "h". (Which just makes "rationalizing the denominator" all that much weirder. It always "unsimplifies" the expression. $\endgroup$
    – fleablood
    Commented Aug 27, 2016 at 22:20

4 Answers 4


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 $$ \lim_{h \to 0}\frac{\sqrt{4+h}-2}{h} = \lim_{h \to 0}\frac{1}{\sqrt{4+h}+2} = \frac{1}{\sqrt{4+0}+2} = \frac{1}{4} $$

  • $\begingroup$ Ah, that makes sense. The wording is a little strange, but I guess I should have focused on the "simplify" part more. Thanks! $\endgroup$ Commented Aug 28, 2016 at 0:04

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!

  • $\begingroup$ True , that is the explanation how derivates work. Why not simply say : rationalize the numerator in order to cancel the $h$ ? This would both give the reader an easy guide what to do and give a good explanation how derivates can be calculated. $\endgroup$
    – Peter
    Commented Aug 27, 2016 at 21:52

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.


$$\frac{\sqrt{4+h}-2}{h}=\frac{\sqrt{4+h}-2}{h}\times \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}=\frac{4+h-4}{h(\sqrt{4+h}+2)}=\frac1{\sqrt{4+h}+2}$$


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