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### Justifying L'Hopital to an ambitious undergrad

Posted: **Wed Sep 18, 2013 3:18 pm UTC**

by **skullturf**

I teach calculus, and I use the classic textbook by Stewart. One topic we cover is L'Hopital's Rule for limits of the form 0/0 or infinity/infinity.

Many students are fine with treating L'Hopital's Rule as a "magic box": For some reason, for the above types of indeterminate form, f/g has the same limit as f'/g', and never mind why.

One of my students is a bit more ambitious, and wants a more thorough reason.

Any suggestions for a nice compelling argument that strikes the right "balance" of rigor and intuition?

Certainly, part of the intuition is that f'(a)/g'(a) can be rewritten as lim_{x->a} ( f(x)-f(a) )/( g(x)-g(a) ). In the relevant section of Stewart, they essentially say: We just consider the case f(a)=g(a)=0, and for other cases, see the appendix. In the appendix, they use Cauchy's Mean Value Theorem, which is perhaps a bit "technical", but maybe that's just life.

Any ideas for a more "cute" or "slick" explanation of L'Hopital that might satisfy a clever and ambitious student?

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Wed Sep 18, 2013 3:53 pm UTC**

by **Qaanol**

An ambitious calc student can definitely handle the MVT. Proving Rolle's theorem is straightforward, and the MVT follows directly from that. Proving LHR is a bit more work, but easily doable.

OTOH if you just want an "intuitive" explanation, differentiability means if you zoom in enough the function is arbitrarily close to a straight line, and the slope of that line is the derivative at that point. Consider two lines through the origin, of different slopes. The ratio of heights of points on those lines equals the ratio of the slopes.

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Wed Sep 18, 2013 8:08 pm UTC**

by **Derek**

Qaanol wrote:OTOH if you just want an "intuitive" explanation, differentiability means if you zoom in enough the function is arbitrarily close to a straight line, and the slope of that line is the derivative at that point. Consider two lines through the origin, of different slopes. The ratio of heights of points on those lines equals the ratio of the slopes.

So I was one of those calculus students who never quite "got" L'Hopital's Rule on an intuitive level, even though I'm very good at math. I knew it was true, and I even knew how to prove it, but it never quite "clicked", until one day I was analyzing a problem, and L'Hopital's Rule came up without me realizing it, and this is the intuition that I used to solve that problem. A few moments later I realized this was a case of L'Hopital's Rule, and I had one of those "eureka" moments.

So yeah, I suggest trying the above. Differentiable functions are locally linear, if two functions both reach zero at the same point, then the ratio of points near the zero will equal the ratio of the slopes (which can easily be proved with actual linear functions, if there is any doubt).

We just consider the case f(a)=g(a)=0, and for other cases, see the appendix. In the appendix, they use Cauchy's Mean Value Theorem, which is perhaps a bit "technical", but maybe that's just life.

Aren't the other cases (f(a) or g(a)=infinity) just handled by using 1/f(a) = 0 (or more rigorously, the limit of 1/f(a) = 0)?

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Wed Sep 18, 2013 8:46 pm UTC**

by **Moole**

I have two proofs that I came up with in my calculus class last year. The first one is pretty intuitive, I think. I like the second one more though.

The statement:

lim x->a f(x)/g(x) = f'(a)/g'(a)

(When the right hand size exists and f(a)=g(a)=0)

is pretty easy to show. Just rewrite the limit as, diving numerator and denominator by (x-a):

lim x->a ((f(x)-f(a))/(x-a))/((g(x)-g(a))/(x-a)) = f'(a)/g'(a)

which is clearly equal and the numerator clearly tends towards f'(a)=lim x->a (f(x)-f(a))/(x-a) and the denominator tends towards g'(a). In fact, you could show that

lim x->a f(x)/g(x) = f''(a)/g''(a)

when f(x), g(x), f'(x), and g'(x) all go to 0 using this (divide by (x-a)^2 instead), and could proceed inductively (though, this technically is a different statement that L'Hopital's rule if all of g's derivatives are 0. Though, I suppose L'Hopital's rule isn't as useful for dealing with such limits anyways).

Another proof (which may be less intuitive) involves the function:

h(x)=xf(x)/g(x)

It's clear that h'(0)=lim x->0 f(x)/g(x), given the limit definition of the derivative. Also, if lim x->0 h'(x) exists, it must equal h'(0). So, examining h'(x) at x other than 0:

h'(x)=(f(x)+xf'(x))/g(x)-(xf(x)g'(x))/(g(x)^2).

and rearranging gives

h'(x)=f(x)/g(x)+(f'(x)/g'(x)-f(x)/g(x))*xg'(x)/x

Then, the limit of that going to zero may be separated out as (assuming all the individual limits exist):

lim x->0 h'(x)=[lim x->0 f(x)/g(x)] + ([lim x->0 f'(x)/g'(x)]-[lim x->0 f(x)/g(x))])*[lim x->0 xg'(x)/x]

and as we have assumed that everything on the right actually exists, the limit lim x->0 h'(x) exists and therefore must equal h'(0). Replacing lim x->0 f(x)/g(x) by h'(0) as well:

h'(0) = h'(0) + ([lim x->0 f'(x)/g'(x)]-[lim x->0 f(x)/g(x))])*[lim x->0 xg'(x)/x].

0 = ([lim x->0 f'(x)/g'(x)]-[lim x->0 f(x)/g(x))])*[lim x->0 xg'(x)/x]

And assuming lim x->0 xg'(x)/x is not 0, this implies

lim x->0 f'(x)/g'(x)=lim x->0 f(x)/g(x)).

(This proof does assume that lim x->0 xg'(x)/g(x) exists and isn't 0, which is kind of an odd condition, but in my proof, I note that "this holds on any g analytic at zero and happens to equal the degree of the lowest non-zero term in the Taylor series expansion at 0", which is pretty easily seen, though I suspect it might actually be the first n such that the nth derivative of g is not zero. )

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Wed Sep 18, 2013 11:15 pm UTC**

by **flownt**

Taylor the function:

lim x->a f(x)/g(x)

=lim x->a (f(a)+(x-a)f'(a)) / (g(a)+(x-a)g'(a))

given that f(a)=g(a)=0:

=lim x->a ((x-a)f'(a)) / ((x-a)g'(a))

=lim x->a (f'(a)) / (g'(a))

Taylor is super useful, also in numerical analysis.

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Thu Sep 19, 2013 6:09 am UTC**

by **cyanyoshi**

I'm fine with L'Hopital's rule. It is great, and it works. It's pretty clear to me why it is true for limits that look like [ lim x->a f(x)/g(x) where f(x) and g(x) approach zero at a], but I never got any kind of intuition of why it should be true for limits as x->infinity or for the other kind on indeterminate form it works on, infinity/infinity. I'm currently a calculus tutor, so this has been bugging me lately more than usual. Telling someone "It works because that's what the textbook says so" and working out a couple of cherry-picked problems just doesn't feel right without at least a vague justification for why L'Hopital's rule works in general.

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Thu Sep 19, 2013 3:31 pm UTC**

by **skullturf**

One thing I tell my students, which is definitely vague and hand-wavy, is along the following lines.

Consider something like 5x/exp(x) as x approaches infinity. The numerator and denominator each approach infinity. However, the numerator grows only linearly, whereas the denominator grows faster than that.

This can be loosely paraphrased by saying the rate of change of the numerator is only constant, but the rate of change of the denominator is increasing.

If instead we started with 5x^2/exp(x), then the numerator is increasing, and the rate of change of the numerator is also increasing, but the rate of change of its rate of change is only constant.

This is all a little vague and I don't necessarily claim that it explains a lot. But it is one thing I briefly mention in an attempt to make L'Hopital slightly more intuitively plausible.

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Sun Sep 22, 2013 9:53 pm UTC**

by **josinalvo**

My intuition so far (and I'll confess it is not complete) is:

1)take two functions f ang g going to infinity. Say that, from a given n, their derivatives are fixed (i.e., f '(x) = c if x>10, g '(x) = d if x>10). After a while the values of f and g are 'dominated' by these derivatives (f(100) = f(10) + c*90, f(1000) = f(10) + c*990)

The ratio becomes (f(10) + c*(x-10)/g(10 + d*(x-10)) which converges to c/d

2)then, we generalize a bit: if, from a given n, the ratio of the derivatives is fixed, we'll get a similar result (instead of c*(x-10) and d*(x-10), we'll have two integrals that differ only by a ratio)

3)then, we remember the definition of limit =P

if the limit of the ratio of the derivatives is c/d (to keep the notation), then there is an n such that the ratio is very near to it. So, appling the argument from 2, we'd know the ratio f/g up to a small constant

### Re: Justifying L'Hopital to an ambitious undergrad

Posted: **Fri Sep 27, 2013 6:07 pm UTC**

by **eSOANEM**

I've never actually proved it, but L'Hopital's never seen unintuitive to me.

I was introduced to big O notation well before I came across L'Hopital's (pretty much the same time I started doing differentiation actually which was about two years or more before I had to treat complicated limits at all properly) and I think that helped.

The way I got my head round it was to think that, usually, for a function f(x)=g(x)/h(x), the limit is O(f(x))=O(g(x)/h(x))=O(g(x))/O(h(x)).

Now, O(f(x)) tells you how quickly f(x) tends towards its limit (or rather the order of convergence or divergence). The derivative also gives you this information albeit in a very different way and so it ought to be obvious that a similar result to the above argument with big-O notation should apply to derivatives. As it turns out, the result is pretty much identical with big Os replaced with derivatives.