The most soluble and miscible proof and the smoothest of logical derivation for a simplified,yet atomic scale interpretation and visualization. Absolutely stupendous!!!
Just as I was trying to understand better L'hopital rule I found your proof, really helped me understand by using the definition of derivative with lim, tysm, wish you the best! :)
thank you for making this the exact information i was looking for, ive watched like 10 different videos on this and they are all too complicated, too fast, or too long to get to the point, your video answered a lot of questions i had that nobody elses videos were covering
I love the proof. It is an ‘ if A, then B’ proof. You start with part of B and jump back to A and use that information to rewrite the expression. When the rewriting from A maths the writing of B, the proof is done. Thank you Doctor. The proof is simple and shiny. Looking forward to the next proof.
I believe in L'Hopital's rule, and I believe in your proof. I am still working on understanding why it makes sense in concept, and I'm almost there. If both numerator and denominator are racing toward infinity, the question is which one gets there faster. In other words, how do their derivatives compare. And since we're heading to infinity, any finite conditions (for example, a constant added to the top or bottom) cease to matter. I think my logic holds up. But when it's 0/0, my logic is a little flimsier. I feel like, if your function is approaching zero, then the reciprocal of your function is approaching infinity, so the same "infinity" logic might apply. But I haven't convinced myself that it's a valid argument.
Thank you so much! this really helped me understand the rule and it's a really elegant proof, and in general your channel is incredible and I cannot believe you don't have more subscribers. However, I've heard that l'hopital's rule works in other cases besides 0/0 like for example infinity*infinity - have I been misinformed or is there some way to further derive other applications of the rule?
Thank you. I hope some day the channels grows sufficiently. Yes it works for any of 'the seven deadly sins'. I have a video of all 7 forms. However, the function must be rewritten as a rational function to apply L"Hospital.
@@PrimeNewtons ah ok, good to know - I actually did watch your seven sins video, so what your saying is that basically all indeterminate forms in some way are derived from 0/0 and as such can have l’hopital’s rule applied to them if expressed as a quotient?
Share a thought? This theorem requires a vivid demonstration for a memory-able understanding. May i suggest the following. Sketch -graph on board: Draw f(x) which is dome -shaped and going through zero at x=a. Also on the same graph, sketch the corresponding f' (x) ; of course with f ' (a)= zero. ..... then also draw th same for a carefully selected g(x).. discuss. what you see. ... Good luck, and have god time having such an enviable job.....suresh
Thank you for this explanation! Can you give us any function which needs another application of l'Hospital's rule ? And by the way your handwriting is nice !
Dear sir. Very Good evening. The explanation part is excellent. The spelling of the rule is to be corrected as I guess. It is L'HOPITAL'S RULE with a hat symbol over O.
Great video! But I miss some explanation about the limit ∞/∞. (and the rule of Hospital is that you go there, when you're ill. You use Hopital for math).
Alright, theres just one important caveat. What if lim x->a f/g is not indeterminate like 0/0? What if it’s defined like 5 or 6. You might think you can use l’hopital anyway. Well it turns out you cannot. The reason is very subtle. If the limit is not indeterminate, then the limit of f/g is the same as when you evaluate f/g at exactly a. We can write the ratio of the derivatives as lim x->a (f(x)-f(a)/g(x)-g(a) )(x-a)/(x-a). The reason that I’m doing this, is that when I evaluate x at a, we get 0/0. This means that we get an undefined result for when we evaluate defined limits. This is quite important to mention.
The only kind of small concern is that for the new limit to be equal to(f(a))'/(g(a))' it probably has yo assume that it is not an indefinite form,but again im not so sure if this is a problem
@@PrimeNewtons I do like the proof. Can you do another video with the rigorous proof? Also one that handles infinity / infinity and the other variations? You do such a magnificent job of presenting, sir! 😃
Better proof and correct oroff involves MEAN VALUE THEOREM : f(x) = f'(x)(x-a), and g(x) =g'(x)(x-a) then a bit of simple algebra yields limit as x approaches a of f'(x)/g'(x) and not simply f'(x)/g'(x).
A really amazing proof But what about the infinity by infinity form😕😕 Also I had a question Say the indertiminate form 1^♾️ For ex a lim g(x) ^f(x) Now say f(a) = 0/0 However it's f'(a) is infinity And g(a) is infinity Then should we apply the standard limit of 1^♾️ form
Mr Newton, you Completely threw me off when you mentioned "checking out" the factors instead of saying "crossing out the factors", which i'm so use to hearing. It has more of a positive connotation when saying checking out the factor !! i think i'll start saying that too. Great video nonetheless.
