$$\lim_{h \rightarrow 0}\frac{5}{\sqrt{5h+1}+1}$$

Some things I'm confused about:

1) Why should we rationalize the denominator? It won't get rid of the square root, it will just move it to the numerator, so the function will still have an irrational part in it. So why do we rationalize the denominator/numerator in these types of functions?

2)This is not a rational function, since the bottom part is not a polynomial and a rational function is composed of polynomials $P$ and $Q$ such that $R(x) = \frac{P(x)}{Q(x)}, Q(x) \neq0.$

3). This function is continuous everywhere where the denominator $\neq0$, correct? The denominator is never equal to zero, right? so is that why this function is continuous and we are able to plug in the value of $h=0$ to evaluate the limit? (we can only plug in values for continuous functions).

I'm just confused on how to solve this problem. Steps & explanations would be very helpful. Thank you.

  • $\begingroup$ There are other problems where it is a good idea to rationalize a denominator, but this isn't one of them. $\endgroup$ – Stefan Smith Nov 1 '13 at 1:13
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    $\begingroup$ The short answer to your question is yes, your point 3) is correct and you can just plug in $h=0$. $\endgroup$ – Stefan Smith Nov 2 '13 at 14:18

Looking at the various comments by OP it seems that the textbooks he is following are not of good quality and instead of providing a step by step approach to limits they are trying to teach a bag of tricks.

For most of the usual limit problems the basic rules of limits and certain standard limits are sufficient:

1) $\displaystyle \lim_{x \to a}f(x) \pm g(x) = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$

2) $\displaystyle \lim_{x \to a}f(x) \cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$

3) $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \dfrac{{\displaystyle \lim_{x \to a}f(x)}}{{\displaystyle \lim_{x \to a}g(x)}}$ provided that $\lim_{x \to a}g(x) \neq 0$.

The above rules apply provided that both limits in RHS exist. Certain standard limits are as follows: $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1},\, \lim_{x \to 0}\frac{\sin x}{x} = 1,\, \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\, \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$

Next we come to the particular question here $$\lim_{h \to 0}\frac{5}{\sqrt{5h + 1} + 1}$$ We don't need to think that this is a rational function or not, but just notice that it is an expression of the form $f(h)/g(h)$ where $f, g$ are some functions and hence the rule 3) should be applied. Clearly for the numerator $f(h) = 5$ we see that $\lim_{h \to 0} 5 = 5$ and we need to see if the limit of denominator $g(h) = \sqrt{5h + 1} + 1$ exists and is non-zero. If we see the form of $g(h)$ it looks like sum of two functions and hence rule 1) can be applied. Thus we can write

$\displaystyle \begin{aligned}\lim_{h \to 0}\frac{5}{\sqrt{5h + 1} + 1} &= \dfrac{{\displaystyle \lim_{h \to 0}5}}{{\displaystyle \lim_{h \to 0}\{\sqrt{5h + 1} + 1\}}}\\ &= \dfrac{5}{{\displaystyle \lim_{h \to 0}\{\sqrt{5h + 1} + 1\}}}\\ &= \dfrac{5}{{\displaystyle \lim_{h \to 0}\sqrt{5h + 1} + \lim_{h \to 0}1}}\\ &= \frac{5}{1 + 1} = \frac{5}{2}\end{aligned}$

Just to provide a contrasting example we try to calculate $$\lim_{h \to 0}\frac{h}{\sqrt{5h + 1} - 1}$$ If we follow as before we will see that limit of both numerator and denominator is $0$ and hence rule 3) can't be applied precisely because the denominator limit is $0$. It is then time to do some manipulation so that the given function can be represented in a form which avoids denominator limit $0$. Such a manipulation is done under the valid assumption that $h \neq 0$. Then

$\displaystyle \begin{aligned}\lim_{h \to 0}\frac{h}{\sqrt{5h + 1} - 1} &= \lim_{h \to 0}\frac{h}{\sqrt{5h + 1} - 1}\cdot\frac{\sqrt{5h + 1} + 1}{\sqrt{5h + 1} + 1}\\ &= \lim_{h \to 0}\frac{h\{\sqrt{5h + 1} + 1\}}{5h}\\ &= \lim_{h \to 0}\frac{\sqrt{5h + 1} + 1}{5}\\ &\text{(following as in previous example)}\\ &= \frac{2}{5}\end{aligned}$

As seen in various limit question on this website, most beginners in calculus try to use concepts like continuity, derivative, L'Hospital and even series expansions to solve simple limit problems (most answers given here also try to use these techniques). It is very unfortunate that beginners underestimate the power of simple rules of limits (mentioned above) and jump on high level concepts. IMHO a beginner studying limits for the first time is better off if he is totally unaware of these high level concepts (which are ultimately derived from simpler concept of limits) and instead focuses on the rules of limits.


Here is a hint: The function $f(h)=\sqrt{5h+1}+1$ is continuous at $x=0$, and $h(0)\neq 0$.


Just about every calculus function is continuous on its entire domain. This includes square roots and many functions containing square roots, such as the one in your question. $0$ is in the domain of your function, so you can compute the limit by "plugging in" 0.

There is no reason to rationalize the denominator.

Stewart's "Calculus" contains the abominable statement that rational functions are continuous on their entire domain. I say "abominable" because it suggests that only rational functions have this property. Are you using Stewart? I don't know how many other texts do this.

EDIT : here are some problems where it is_ a good idea to rational a denominator , and may be necessary :

Simplify $\frac{1}{1+\sqrt{3}}+\frac{2}{5-\sqrt{3}}$

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

  • $\begingroup$ why don't we rationalize the denominator? My text says to rationalize it. In which cases do we rationalize numerators/denominators? Thank you. $\endgroup$ – Emi Matro Nov 1 '13 at 1:14
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    $\begingroup$ In this case, don't. It contributes nothing to solving the problem. $\endgroup$ – ncmathsadist Nov 1 '13 at 1:16
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    $\begingroup$ Your text is giving you terrible advice . what text is it ? rationalizing the denominator is not only unnecessary , it is counterproductive, because then the denominator will tend to zero as $h$ goes to zero ,making the problem moe confusing. You should show your teacher our objections and recommend changing textbooks . This is even worse than Stewart 's gaffe. $\endgroup$ – Stefan Smith Nov 1 '13 at 2:50
  • $\begingroup$ Please see addition up my answer $\endgroup$ – Stefan Smith Nov 1 '13 at 3:04

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