L'Hôpital's rule review (article) | Khan Academy (2024)
L'Hôpital's rule helps us find many limits where direct substitution ends with the indeterminate forms 0/0 or ∞/∞. Review how (and when) it's applied.
Want to join the conversation?
Log in
∆
4 years agoPosted 4 years ago. Direct link to ∆'s post “Why is 1^infinity an inde...”
Why is 1^infinity an indeterminate form?
•
(4 votes)
The #1 Pokemon Proponent
4 years agoPosted 4 years ago. Direct link to The #1 Pokemon Proponent's post “This stems from the fact ...”
This stems from the fact that all of the limits in calculus of this type have something to do with the number e (2.71828...)
e is actually defined as limit(n->infinity, (1+1/n)^n). At first glance, (1+1/n) seems to be 1 and hence, this is called 1^infinity form. However, the limit of this quantity is 2.718...
(5 votes)
Yash
6 years agoPosted 6 years ago. Direct link to Yash's post “In the article's example ...”
In the article's example for using L'Hopital's rule for finding limits of exponents, they get (1+2(0))^1/sin(0) = 1^infinity (direct substitution). But won't 1/sin(0) be undefined, thus resulting in 1^undefined = undefined?
could someone help? the question was to find the limit as x approaches 1 from the positive side for (x-1)^(x-1)
•
(1 vote)
Peter Ryner
2 months agoPosted 2 months ago. Direct link to Peter Ryner's post “(x-1) is the same as 1/(x...”
(x-1) is the same as 1/(x-1)^-1. Putting it in that form makes it useful for checking L'Hopital's Rule because we don't care about the overall function/quotient to start with, just the individual functions themselves.
The limit of (x-1)^-1 = 1/(x-1) as x approaches 1 from the positive direction is infinity, which is readily apparent if you graph it on Desmos or some such. Going through L'Hopital's rule you'll eventually get the limit of Ln(y) = 0, so for that to be true the limit of y, and thus the limit we want, must be 1. Hope that helps.
8 years agoPosted 8 years ago. Direct link to hummusw's post “Is there a printable vers...”
Is there a printable version of this page?
•
(1 vote)
ondraperny
7 years agoPosted 7 years ago. Direct link to ondraperny's post “Unfortunately there is no...”
Unfortunately there is nothing like that. So far best solution might be using Snipping tool on windows which can easily cut "pictures" from browser and then you can arrange them together in some software( Microsoft word would suffice).
(1 vote)
Osmis
2 years agoPosted 2 years ago. Direct link to Osmis's post “I had a problem (1-4/x)^x...”
I had a problem (1-4/x)^x . My question was when they took (4/x^2)/((1-4/x)(-1/x^2)) and got (4x^2)/(1-4/x)(-x^-2).
•
(1 vote)
Vincent Pace
7 years agoPosted 7 years ago. Direct link to Vincent Pace's post “When using L'Hôpital's ru...”
When using L'Hôpital's rule to find limits of exponents, there's a step that sets, for example, lim x->∞ ln(y) equal to ln (lim x->∞ y). Which logarithm or limit property allows this?
•
(1 vote)
Paras Sharma
7 years agoPosted 7 years ago. Direct link to Paras Sharma's post “Here we can use this prop...”
Here we can use this property because here we are not applying the limit to whole ln(y(x)) operator we have our variable x in the y(x) , So here we just wanna find the limiting value of y(x) It doesn't violate our previous method that we use we just plug the value and try to come up w/ a more subtle and concrete way of understanding this.
(1 vote)
John He
7 years agoPosted 7 years ago. Direct link to John He's post “What about lim x→0 cot(x)...”
What about lim x→0 cot(x)/In(x)?If you apply L'Hôpital's rule,try to differentiate this,you will get into great trouble!
•
(0 votes)
kubleeka
7 years agoPosted 7 years ago. Direct link to kubleeka's post “Direct substitution gives...”
Direct substitution gives ∞/∞, so taking the derivatives of according to l'Hôpital yields -csc²(x)/(1/x). This rearranges into -x/sin²(x). Direct substitution now yields 0/0, so we can apply l'Hôpital's Rule again. Differentiate to get -1/(2sin(x)cos(x))
Now, finally, direct substitution yields -1/0, which indicates that the limit does not exist.
What is L'Hôpital's rule? L'Hôpital's rule helps us evaluate indeterminate limits of the form or . In other words, it helps us find lim x → c u ( x ) v ( x ) , where lim x → c u ( x ) = lim x → c v ( x ) = 0 (or, alternatively, where both limits are ).
L'Hospital's rule is a general method of evaluating indeterminate forms such as 0/0 or ∞/∞. To evaluate the limits of indeterminate forms for the derivatives in calculus, L'Hospital's rule is used. L Hospital rule can be applied more than once.
Quick Overview. Recall that L'Hôpital's Rule is used with indeterminate limits that have the form 00 or ∞∞. It doesn't solve all limits. Sometimes, even repeated applications of the rule doesn't help us find the limit value.
L' Hospital's rule states that, when the limit of f(x)g(x) is indeterminate, under certain conditions it can be obtained by evaluating the limit of the quotient of the derivatives of f and g (i.e., f′(x)g′(x)). If this result is indeterminate, the procedure can be repeated.
Continuous compounding interest rates encountered everyday in investments, different types of bank accounts, or when paying credit cards bills, mortgages, etc. In short, L'Hospital's rule has many applications in the real world, predominantly in statistics, physics, and engineering!
Indeterminate forms besides "0"0 and "∞"∞ include "0⋅∞, "∞−∞", "1∞", "00", and "∞0". These forms also arise in the computation of limits and can often be algebraically transformed into the form "0"0 or "∞"∞, so that l'Hopital's Rule can be applied.
L'Hô pital's Rule can be strengthened to include the case when g′(a)=0 and the indeterminate form " ∞/∞ ", the case when both f and g increase without any bound.
Sometimes we will need to apply L'Hospital's Rule more than once. L'Hospital's Rule works great on the two indeterminate forms 0/0 and ±∞/±∞ ± ∞ / ± ∞ . However, there are many more indeterminate forms out there as we saw earlier.
Note that both x and e^x approach infinity as x approaches infinity, so we can use l'Hôpital's Rule. Also, the derivative of x is 1, and the derivative of e^x is (still) e^x.
L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives. Find the derivative of the numerator and denominator. Differentiate the numerator and denominator.
The assumptions of L'Hospital's rule are that the functions f and g are differentiable, that they are an indeterminate form of type 0/0 or infinity/infinity, and that g'(x) is not zero near the limit point. It is also required that the limit of f'/g' as x approaches a exists.
When Can You Use L'hopital's Rule. We can apply L'Hopital's rule, also commonly spelled L'Hospital's rule, whenever direct substitution of a limit yields an indeterminate form. This means that the limit of a quotient of functions (i.e., an algebraic fraction) is equal to the limit of their derivatives.
Do not apply L'Hopital's rule if the limit is not resulting in an indeterminate form. We can apply L'Hopital's rule as many times as required but before the application of each time, we should check whether the limit in that particular step is giving indeterminate form.
L'Hôpital's rule can be used to evaluate the limit of a quotient when the indeterminate form 0/0 or ∞/∞ arises. L'Hôpital's rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form 0/0 or ∞/∞ .
The tangent to the curve at the point [g(t), f(t)] is given by [g′(t), f′(t)]. L'Hôpital's rule then states that the slope of the curve when t = c is the limit of the slope of the tangent to the curve as the curve approaches the origin, provided that this is defined.
If f(x) and g(x) are continuous functions with either f(a)=g(a)=0, or f(a)=±∞ and g(a)=±∞, then limx→af(x)g(x)=limx→af′(x)g′(x). Warning: While L'Hospital's Rule says that the limits of fg and f′g′ are equal under certain circ*mstances, it is not true that fg=f′g′!
The letter L is used to represent the limit in calculus because it is a traditional notation that has been used for centuries. The notation for limit was first introduced by the mathematician Augustin-Louis Cauchy in 1821, and it has been widely used ever since.
Introduction: My name is Nathanael Baumbach, I am a fantastic, nice, victorious, brave, healthy, cute, glorious person who loves writing and wants to share my knowledge and understanding with you.
We notice you're using an ad blocker
Without advertising income, we can't keep making this site awesome for you.
