# Why does L'Hospital's rule work?

Can anyone tell me why the L'Hospital's rule works for evaluating limits of the form $\frac{0}{0}$ and $\frac{\infty}{\infty}$ ?

What I understand about limits is that when you divide a really small thing (that is $\rightarrow0$) by another really small thing, we get a finite value which may not be so small.

So how does differentiating the numerator and denominator help us get the Limit of a function?

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This question serves a great deal for undergraduate students who get the same doubt when they are reading the principle. – Iyengar Jan 11 '12 at 17:12

The answer given by Mr. Jackson Walters gives a proof why it works, but if you are looking for an answer that should give an intuition, see this :

Consider the curve in the plane whose $x$-coordinate is given by $g(t)$ and whose $y$-coordinate is given by $f(t)$, i.e. $$\large t\mapsto [g(t),f(t)].$$ Suppose $f(c) = g(c) = 0$. The limit of the ratio $\large \frac {f(t)}{g(t)}$ as $t \mapsto c$ is the slope of tangent to the curve at the point $[0, 0]$. The tangent to the curve at the point $t$ is given by $[g'(t), f'(t)]$. l'Hôpital's rule then states that the slope of the tangent at $0$ is the limit of the slopes of tangents at the points approaching zero.

Points to assume (credits : Thanks to Hans lundmark for pointing out what I missed and to Srivatsan for improving my formatting . )

Assume that functions $f$ and $g$ have a well defined Taylor expansion at $a$.

Proof:

Another way you can think of this is to use the idea of derivative: a function $f(x)$ is differentiable at $x=a$ if $f(x)$ is very close to its tangent line $y = f'(a) \cdot (x-a) + f(a)$ near $x = a$. Specifically,

$$f(x) = f(a) + f'(a) \cdot (x-a) + E_{1}(x)$$

where $E_{1}(x)$ is an error term which goes to $0$ as $x$ goes to $a$. In fact, $E_{1}(x)$ must approach $0$ so fast that

$$\lim_{x\to a}\frac{E_1(x)}{x-a}=0$$

because $\dfrac{E_{{1}(x)}}{x-a} = \dfrac{ f(x)-f(a) }{x-a} - f'(a)$

and we know from the definition of derivative that this quantity has the limit $0$ at $a$.

Similarly, if $g$ is differentiable at $x = a$,

$$g(x) = g(a) + g'(a) \cdot (x-a) + E_{2}(x)$$

where $E_{2}(x)$ is another error term which goes to $0$ as $x \to a$. If you're computing the limit of $f(x)/g(x)$ as $x \to a$ and if $g(a)$ is not equal to $0$, then as $x \to a$, the numerator becomes indistinguishable from $f(a)$ and the denominator from $g(a)$, so the limit is

$$\lim_{x \to a} \frac{f(x)}{g(x)}=\frac{f(a)}{g(a)} .$$

If both $f(a)$ and $g(a)$ are $0$, then we must use the tangent approximations to say that

$$\frac{f(x)}{g(x)} = \frac{f(a) + f'(a) \cdot (x-a) + E_{1}(x) }{ g(a) + g'(a) \cdot (x-a) + E_{2}(x) }$$

$$=\frac{f'(a) \cdot (x-a) + E_{1}(x)}{g'(a) \cdot (x-a) + E_{2}(x) }$$

$$=\frac{f'(a) + [E_{1}(x)/(x-a)] }{g'(a) + [E_{2}(x)/(x-a)]}$$

and we have seen that the second term becomes negligible as $x\mapsto a$.

In other words, when both function values approach $0$ as $x\mapsto a$, the ratio of the function values just reduces to the ratio of the slopes of the tangents, because both functions are very close to their tangent lines.

I hope you understood. Thanks a lot. Iyengar.

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Same comment here as for Jackson's answer: This assumes that $f$ and $g$ have Taylor expansions at $a$, which is a stronger assumption than necessary. (But of course it's still a good argument for giving a bit of intuition.) – Hans Lundmark Jan 11 '12 at 11:28
Do you want a complete argument ?, ok let me post it @HansLundmark – Iyengar Jan 11 '12 at 12:23
@iyengar: I have cleaned up the TeX in your post a bit, hope it is ok. // In case you are interested, these are the changes I made: (1.) I used the pre-defined operator \lim. (2.) Though the meaning is quite clear, I have not seen "$x$ approaches $a$" written as $x \mapsto a$; I changed it to x \to a. (3.) The derivative of $f$ can be typeset as f^\prime or f'. f^' makes the "dash" symbol too small. – Srivatsan Jan 11 '12 at 15:52
@Srivatsan : Infinitely many thanks to what you have done sir, I am in debt with you for what you have done, Thanks a lot !!, you have all rights to make any changes sir, you never need to give explanations to me, I am a humble student before you. – Iyengar Jan 11 '12 at 16:42
@iyengar: The point of the comment is two-fold: (1.) It is common courtesy to inform the answerer of extensive typesetting/content changes, such as the one I made. (2.) I wanted to explain the changes I made so that you can learn them and implement them in your future posts. In case you would like any clarifications in any of the edits, please feel free to ask. // On a side note, I liked this answer, thanks. – Srivatsan Jan 11 '12 at 17:02

This is far from rigorous, but the way I like to think about L'Hospital's Rule is this:

If I have a fraction whose numerator and denominator are both going to, say, infinity, then I can't say much about the limit of the fraction. The limit could be anything.

It's possible, though, that the numerator goes slowly to infinity and the denominator goes quickly to infinity. That would be good information to know, because then I would know that the denominator's behavior is the one that really swings the limit of the fraction overall.

So, how can I get information about the rate of change of a function? This is precisely the kind of thing a derivative can tell you. Thus, instead of comparing the numerator and denominator directly, I can compare the rate of change (i.e. the derivative) of the numerator to the rate of change (i.e. the derivative) of the denominator to determine the limit of the fraction overall. This is L'Hospital's Rule.

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Thank you for explaining it in a language that normal people can understand. – Dimme Jan 14 '12 at 0:52
@Dimme You're welcome, but what I've written is not an explanation, per se. At best, I've indicated why L'Hospital's Rule ought to work. The more thorough answers in this thread prove that it does work. – Austin Mohr Jan 14 '12 at 1:26
Understanding the problem is half the solution. – Dimme Jan 14 '12 at 3:58
This is exactly the way I like to think about L'Hospital's rule! +1 – M Turgeon Mar 9 '12 at 17:49

For the case $\frac{0}{0}$, we need only to use definition of a derivative in terms of the difference quotient. Suppose $f,g:\mathbb{R} \rightarrow \mathbb{R}$ and $f(a)=g(a)=0$, $f$ and $g$ are continuously differentiable, and $g'(a) \ne 0$, then

$$\begin{eqnarray*} \lim_{x \rightarrow a}\frac{f(x)}{g(x)} &=& \lim_{x \rightarrow a}\frac{f(x)-0}{g(x)-0} \\ &=&\lim_{x \rightarrow a}\frac{f(x)-f(a)}{g(x)-g(a)}\\ &=&\lim_{x \rightarrow a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}\\ &=&\frac{\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}}{\lim_{x \rightarrow a}\frac{g(x)-g(a)}{x-a}}\\ &=&\frac{f'(a)}{g'(a)} \end{eqnarray*}$$

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This is not entirely correct. The last step should be $f'(a)/g'(a)$ instead of $\lim_{x\to a} f'(x)/g'(x)$. These two need not be equal in general, so your computation only proves the rule under the extra assumptions that $f$ and $g$ are continuously differentiable at $a$, and that $g'(a) \neq 0$. It holds under more general assumptions, but then the proof is more complicated. – Hans Lundmark Jan 11 '12 at 7:53
Good point, I added the necessary quantifiers. – Jackson Jan 11 '12 at 8:03

At the heart of it though, L'Hopital's rule just seems to be a marriage of the ideas that differentiable functions are pretty darn close to their linear approximations at some point as long as you don't stray too far from that point and that for a continuous function, a small movement in the domain means a small movement in the value of the function.

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