Today a student in my calculus class asked why L’Hopital’s Rule works. I paused, and failed to think of an easy way to explain it. But now I’ve found a really easy way to explain it — at least for the 0/0 case. (Thanks Rogawski!)

We want to show that \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a} f'(x)}{\lim_{x\rightarrow a} g'(x)}.

At least in the 0/0 case, we know that f(a)=0 and g(a)=0. Great! If that be true, we can say that:


Of course that has to be true, because we’re subtracting 0 from the top and the bottom! Now we can say:


(We are dividing the top and bottom by the same number.)

Finally, we take the limit as x approaches a of both sides, to get:

\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}

By basic limit rules, we can rewrite the right hand side of the equation to be the limit of the top and bottom separately. But the limit of the top and bottom separately are just the derivatives! (See the definition of the derivative there?)

\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a} f'(x)}{\lim_{x\rightarrow a} g'(x)}




  1. Nifty!

    This is what popped into my head after reading your first sentence (and pausing to see how I’d answer it). I don’t know if this is too hand-wave-y, but how about this… if f(a)=g(a)=0 (and their first derivatives exist), then as x->a the functions can be approximated by f(x)=f'(a)(x-a) and g(x)=g'(a)(x-a). Then putting f(x)/g(x), the (x-a) terms cancel, giving you f'(a)/g'(a). Or something.

  2. @Matt: Ah, this is how our textbook (Anton) does it. I couldn’t show that to my regular calc students, because we don’t talk about function approximations in that way. (Although I know you’re just finding the equation to the line tangent to the function at a point. Which they know how to do.)

    But I don’t think it’s handwavy at all. In fact, your way really does explain things neatly.

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