But Why Does L'Hopital's Rule Work?

[@lorenzopirozzi7298]

It is important to remember that the limit of the derivatives is equal to the original limit provided that it exists. There are quite a few cases in which applying L’H gives you a limit that doesn’t exist, in contrast to the starting limit (if I am not mistaken one example of that is the limit of x->inf of x+sinx/x-sinx). In my experience, L’H is a good tool, but it requires quite a bit of calculations so usually it’s just better to use Taylor series expansions. Great video nonetheless!

[@Vannishn]

Very nice ! Thank you :) I think your assumptions on f and g are that they are both equal to 0 and differentiable around c. For the case ±infinty, the sign can be factored, and then you can flip top and bottom by taking the inverse, which then each indeed tend to 0.

[@michaurbanski5961]

Very cool, especially for the 0/0 case! For the 8/8 case, can we show that the gradient always takes over? It seems hard to prove for all functions, it even seems false (are there no functions that take over their gradients?)

[@ritvikmath]

Here's an interesting way to think about it! We know that the function e^x has the property that its gradient is itself e^x. In other words, e^x exactly keeps pace with its own gradient. But e^x doesn't "grow fast enough" to have a vertical asymptote. So, in some sense, any function which grows fast enough to have a vertical asymptote must do so by growing an order of magnitude faster than its own value. Not a proof, I'll admit, but it helped me justify this idea in my own head.

[@michaurbanski5961]

@@ritvikmath Nice, thanks!

[@enricolucarelli816]

For the 8/8 limit, I have come up to a, in my opinion, nicer proof: f / g = (1/g) / (1/f) thus we go back to a 0/0 limit. Aplying L’H we get L = L^2 lim g’/f’ deducing that L = lim f’/g’

[@michaurbanski5961]

@@enricolucarelli816 I don't think it works, since (1/f)' is not (1/f') So even though we can make 8/8 into 0/0, we would get different derivatives

[@enricolucarelli816]

@ I know! But if you do it carefully you’ll see it works out! (1/g(x))’= - g’/g^2 and (1/f(x))’= - f’/f^2 Substituting you get: L = L^2 lim g’/f’ -> 1/L = lim g’/f’ -> L = lim f’/g’

[@martinkunev9911]

13:10 The derivatives are being multiplied by a very small number h, which in the limit approaches 0. It is not obvious that the left terms are being outpaced.

[@suhail_69]

Great point !!

[@ritvikmath]

yes that's an awesome observation and I'm gonna need to think more on it! I think it's a great highlight of what gets missed when trying to condense a formal mathematical proof into an "easier to understand" pseudo-proof

[@snnehal]

😄 nice tie in to the life lesson

[@ritvikmath]

😂 thanks

[@deleted-something]

thanks!

[@frankansari3457]

I remember this rule back from school - but I admit I have not thought about this for decades. 😂

[@ritvikmath]

😂