Log in Sarah 10 years agoPosted 10 years ago. Direct link to Sarah's post “So I've watched all of th...” So I've watched all of the videos on L'Hoptial's Rule but I'm still unsure of how to deal with the other indeterminate forms. In class, my teacher would take the natural log of the functions and then solve but I was so confused when we learned that. If anybody knows what I'm talking about could you please help me? • (26 votes) Just Keith 10 years agoPosted 10 years ago. Direct link to Just Keith's post “For other indeterminate f...” For other indeterminate forms, you need to do mathematical manipulations to get the limit into a l'Hôpital form -- 0/0 or infinity/infinity. Sometimes, that can be done, sometimes it cannot. If you cannot get it into a l'Hôpital form, you cannot solve the limit this way. Here is what to do (which won't always work) for common indeterminate forms: Infinity - infinity: Find the common denominator and combine the two expressions into a single fraction. If that gives you 0/0 or infinity/infinity then you can proceed with l'Hôpital's Many of the other indeterminate forms that involve exponents can sometimes be put into l'Hôpital forms by doing the following: Put the entire limit to either e^ln(the limit) or ln(e^limit), whichever works better (more often, it will be the first form). You may legitimately move the lim x→ ? between the e and the ln. Thus, The reason that this method works is because e^ and ln are inverse functions and cancel each other out. Of course, this won't always work and you sometimes have to use other methods. Sometimes, you cannot solve the limit at all (though you probably won't be given one of those). (46 votes) Tristin Solorzano 10 years agoPosted 10 years ago. Direct link to Tristin Solorzano's post “How hard is it to prove t...” How hard is it to prove the general case? • (8 votes) Laserbeamcool 10 years agoPosted 10 years ago. Direct link to Laserbeamcool's post “It gets quite involved, u...” It gets quite involved, using like Cauchy's Mean Value Theorem and Rolle's Theorem etc. 18 minute proof on Youtube http://www.youtube.com/watch?v=FhFWfVUBXC4 (11 votes) euler 10 years agoPosted 10 years ago. Direct link to euler's post “Since 0^0 is indeterminat...” Since 0^0 is indeterminate, how do we apply l'Hopital's rule? Thanks in advance • (4 votes) Just Keith 10 years agoPosted 10 years ago. Direct link to Just Keith's post “You have to convert it in...” You have to convert it into either a 0/0 or ∞/∞ form or some other form you know how to take the limit of. Exactly how you do that depends on the details of the problem. Let us do a simple example (6 votes) Chirag Vasanth 10 years agoPosted 10 years ago. Direct link to Chirag Vasanth's post “Could you please tell me ...” Could you please tell me where i can get the complete proof of L'Hospital's rule?? A video explanation would be preferable. • (5 votes) Physics and Poets 8 years agoPosted 8 years ago. Direct link to Physics and Poets's post “I have seen al these L'Ho...” I have seen al these L'Hopital 's Rule videos and yet, I still have a question. This proof video was highly helpful, but it only addressed when the function behaved so that it ended up "0/0" Zero over Zero, where the top and bottom functions both went towards zero. Could someone please post the proof if the 2 functions behaved: Infinity / Infinity? How would you prove from that route? What I am asking is instead of what Sal did, where f(a) = 0 and g(a) = 0, prove it with f(a) = infinity and g(a) = infinity. Please respond ASAP. Thanks in Advance • (4 votes) Mark Regan 4 years agoPosted 4 years ago. Direct link to Mark Regan's post “Dear Physics and Poets,I...” Dear Physics and Poets, (2 votes) mahmoud 11 years agoPosted 11 years ago. Direct link to mahmoud's post “how can i solve the l'hop...” how can i solve the l'hopital rule in case the limits appr to infinity ? • (2 votes) Scout McComb 11 years agoPosted 11 years ago. Direct link to Scout McComb's post “Keep taking the derivativ...” Keep taking the derivative until you are left with a constant. (3 votes) Hyeon Myeong Jeong 11 years agoPosted 11 years ago. Direct link to Hyeon Myeong Jeong's post “Well, L'Hôpital's(=L'Hosp...” Well, L'Hôpital's(=L'Hospital's) Rule, in special case, this proof is so simple. but, Is it work if one of 2 functions, either f(x) or g(x), is radical, exponential, logarithmic, or trigonometric functions? • (2 votes) Sahil Acharya 11 years agoPosted 11 years ago. Direct link to Sahil Acharya's post “yes, as long as fx and gx...” yes, as long as fx and gx both approach 0 or +- infinity, then lhopitals rule can be used. the type of function doesnt really matter (3 votes) preethvijay17 10 years agoPosted 10 years ago. Direct link to preethvijay17's post “At 2:26, when sal is rewr...” At • (1 vote) Creeksider 10 years agoPosted 10 years ago. Direct link to Creeksider's post “There's more than one way...” There's more than one way to write the formula for calculating the derivative, and in this case it's merely a matter of personal preference which one you choose. Suppose we got tired of writing x + delta x, and create a new variable "a" equal to x + delta x. Now the top part of the fraction is f(a) - f(x). The bottom part is now a - x instead of delta x. You can easily see that if a = x + delta x, then delta x = a - x. There's one more change we have to make. Instead of taking the limit as delta x goes to zero, we take the limit as a goes to x. Again, it's easy to see that if a = x + delta x, these two things are equivalent, because when "a" gets closer and closer to x, delta x (the difference between them) gets closer and closer to zero. (4 votes) akibshahjahan 11 years agoPosted 11 years ago. Direct link to akibshahjahan's post “at 1:30, it looks like th...” at • (0 votes) doctorfoxphd 11 years agoPosted 11 years ago. Direct link to doctorfoxphd's post “Yup, it is the same rule,...” Yup, it is the same rule, but ONLY PART OF IT, just a subset of all the conditions that are covered by L'Hôpital's Rule. That is his purpose with this video, providing a In terms of the context, well, sometimes you are on a test or assignment and you cannot proceed because the simplest way of getting to a result is through one of the indeterminate combinations like 0/0. When you encounter that, you can either give up and start head-butting the nearest wall, or you think (6 votes) Raquel Monforte 4 years agoPosted 4 years ago. Direct link to Raquel Monforte's post “I'm sorry I dont understa...” I'm sorry I dont understand how this case is different to the other ones? Or is he just proving for when we use for 0/0 and it doesn't apply to infinity/infinity? I just dont understand why they call it a special case when that is just the general rule? • (2 votes)Want to join the conversation?
you can get (for the e^ln(original limit)
e^[lim x→ ? {ln {original function}}
You can then use the properties of ln to move the exponent down, as in this form:
e^[lim x→ ? (original exponent){ln {base of the original function}}
Now, looking only at:
lim x→ ? (original exponent){ln {base of the original function}
You can take the limit -- either directly, if that is possible, or via l'Hôpital's if you have 0/0 or infinity/infinity.
After you take that limit, remember that your final answer is:
e^(whatever that limit was).
y = lim x→0 of x^x
ln y = lim x→0 x ln x
Right hand side is now a 0 ∙ ∞ form. Still needs more manipulation. Remember that x = 1/(1/x) so we apply that:
ln y = lim x→0 ln x / (1/x)
The Right hand side is now ∞/∞ form, so we can use L’Hˆopital’s
ln y = lim x→0 (1/x) / (-1/x²)
(1/x) / (-1/x²) simplifies to −x, thus
ln y = lim x→0 −x
ln y = 0
But this is not the original form, but rather the ln of the original. We need to reverse that:
e^(ln y) = e^0
y = 1
Thus, lim x→0 x^x = 1
I am as ignorant as you regarding how to prove it using f(a) and g(a) --> infinity. But I AM LOOKING
INTO IT AND WILL REPLY SOON. Today is 04/19/2020. Reply sooner if you have figured out already please,
Mark Regan
2:26 1:30
i dont seem to see the special case...
also i dont understand the context of this rule and the proof
What is special is that this subset of the rule (when f(a) = 0 and g(a) =0 , and the derivatives both exist) IS that it supports a simple and straightforward proof. proof of this subset of L'Hôpital's Rule.
In showing the proof, he gave us a nice example of the proof process and a nice trick of constructing a proof by proving it backwards.L'Hôpital and happily get your result. Or, maybe you need to provide a proof, and you can think of this nifty trick and succeed to the cheers of thousands.
Video transcript
What I want to goover in this video is a special caseof L'Hopital's Rule. And it's a moreconstrained version of the general casewe've been looking at. But it's still very powerfuland very applicable. And the reason why we're goingto go over this special case is because its proof isfairly straightforward and will give you an intuitionfor why L'Hopital's Rule works at all. So the special caseof L'Hopital's Rule is a situation wheref of a is equal to 0. f prime of a exists. g of a is equal to 0. g prime of a exists. If these constraintsare met, then the limit, as x approaches a off of x over g of x, is going to be equal to fprime of a over g prime of a. So it's very similarto the general case. It's little bitmore constrained. We're assuming thatf prime of a exists. We're not justtaking the limit now. We're assuming f prime of a andg prime of a actually exist. But notice if we substitutea right over here we get 0/0. But that if thederivatives exist we could just evaluatethe derivatives at a, and then we get the limit. So this is very closeto the general case of L'Hopital's Rule. Now let's actually prove it. And to prove it, we're goingto start with the right hand and then show that if we usethe definition of derivatives, we get the left handright over here. So let me do that. So I'll do it right over here. So f prime of ais equal to what, by the definitionof derivatives? Well, we could viewthat as the limit as x approaches a of f of xminus f of a over x minus a. So this is literally justa slope between two points. So like, if you have yourfunction f of x like this, this is the point a, fof a right over here. This right over hereis the point x, f of x. This expressionright over here is the slope betweenthese two points. The change in our y valueis f of x minus f of a. The change in our fvalue is x minus a. So this expression is justthe slope of this line. And we're just takingthe-- let me actually do that in a differentcolor-- the line that connects these two points,that's the slope of it. I'll do that in white. The slope of the line thatconnects those two points. And we're takingthe limit as x gets closer and closerand closer to a. So this is justanother way of writing the definition ofthe derivative. So that's fine. Let's do the samething for g prime of a. So f prime of aover g prime of a, is going to be thisbusiness which is in orange, f prime of a over g prime of a. Which we can write asthe limit as x approaches a of g of x minus gof a over x minus a. Well, in the numerator,we're taking the limit as x approaches a, andin the denominator, we're taking the limitas x approaches a. So we can just rewrite this. This we can rewrite asthe limit as x approaches a of all thisbusiness in orange. f of x minus f ofa, over x minus a, over all the business in green. g of x minus g of a, allof that over x minus a. Now, to simplify this, wecan multiply the numerator and the denominator by x minusa to get rid of these x minus a's. So let's do that. Let's multiply by xminus a over x minus a. So the numerator, x minus a,and we're dividing by x minus a. Those cancel out. And then these two cancel out. And we're left with this thingover here is equal to the limit as x approaches aof, in the numerator we have f of x minus f of a. And in the denominator, wehave g of x minus g of a. And I think you seewhere this is going. What is f(a) equal to? Well, we assumed fof a is equal to 0. That's why we'reusing L'Hopital's Rule from the get go. f of a is equalto 0, g of a is equal to 0. f of a is equal to 0.g of a is equal to 0. And this simplifies to thelimit as x approaches a of f prime of x, sorry of f ofx, we've got to be careful. Of f of x over g of x. So we just showed that if f ofa equals 0, g of a equals 0, and these two derivatives exist,then the derivatives evaluated at a over each other aregoing to be equal to the limit as x approaches a off of x over g of x. Or the limit as xapproaches a of f of x over g of x is going tobe equal to f prime of a over g prime of a. So fairly straightforwardproof for the special case-- the special case, notthe more general case-- of L'Hopital's Rule.
