Using L'Hopital's Rule to show that exponentials dominate polynomials
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- Опубліковано 10 лис 2024
- Description:
L'Hopital's rule tells us that the limit of a quotient f(x)/g(x) of "indeterminant form" is the same as the limit for f'(x)/g'(x). Bizarre! Nonetheless, we can use this useful tool to show that e^x dominates the growth of x, x^2, or any polynomial.
Learning Objectives:
1) Use L'Hopital's Rule to compute the limit of functions in the indeterminate forms 0/0, infinity/infinity and infinity - infinity.
Now it's your turn:
1) Summarize the big idea of this video in your own words
2) Write down anything you are unsure about to think about later
3) What questions for the future do you have? Where are we going with this content?
4) Can you come up with your own sample test problem on this material? Solve it!
Learning mathematics is best done by actually DOING mathematics. A video like this can only ever be a starting point. I might show you the basic ideas, definitions, formulas, and examples, but to truly master calculus means that you have to spend time - a lot of time! - sitting down and trying problems yourself, asking questions, and thinking about mathematics. So before you go on to the next video, pause and go THINK.
This video is part of a Calculus course taught by Dr. Trefor Bazett at the University of Cincinnati.
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THIS IS SO HELPFUL TY!! you’re wonderful at explaining things
So good, sir. Request for Cauchy's Mean Value Theorem for L'Hopital's Rule.
Very good video. I have some constructive criticism, if you'd like. Your 2s could be confused with alphas or the partial derivative symbol. And at 7:20, I would have given the guess for the correct result ("e^x ultimately grows faster than x^2, so the limit should approach infinity, but can we prove it with L'Hopital's Rule?") and then done the algebra trick. But I really like your tone, and the visual setup, and the way you progress, and everything really. Your channel will help a lot with my study. Thanks.
Dr. Bazett, you have 40 (shapes of graphs) and 39 (1st and 2nd derivative test) reversed above; they should be 39 and 40, respectively, I think. I'm loving this course.
Interesting and fun
Love from punjab india
Hi prof Trefor, what's the intuition behind L'Hopital's rule ? I could understand taking the 1st derivative, but can't see the intuition behind taking the 2nd, 3rd.... derivative when the numerator/denominator functions still compute to 0/0 or infinity/infinity. Another thing is that f(x)/g(x) at a specific x value takes the quotient of f evaluated at x and g evaluated at x, however, take the example of f(x) and g(x) crossing each other at x=a, where f & g become 0, the quotient of f/g at the left hand side of the intersection is almost the inverse of the f/g at the right hand side of the intersection, when lim x->a from both sides, yet L'Hopital's rule takes the derivative of f and g with lim->a making the quotient of f'/g' regardless if it's evaluated from the left or right of x=a, which is somewhat counter-intuitive.
7:05 , I will not talk of millions or trillions of power but anything above cubic function should eventually dominate in the long run. X^3 , X^4 anything above dominates. Please explain why limit is still converging to zero.?
You can easily verify that e^x dominates x^4 by plugging in x = 10.
As for why, the reason is that we can apply L'Hopital's rule as long as we have the limit evaluating to inf/inf or 0/0. When we plug in infinity into x^4/e^x we get inf/inf. Therefore, we can apply L'Hopital's rule to this limit. When we do that, we get 4*x^3/e^x. Again, plug in infinity, you get inf/inf. Apply L'Hopital's rule again. 12*x^2/e^x. As per limit laws, the 12 can be taken out of the limit (that is, lim(12 * x^2/e^x) = 12 * lim(x^2/e^x)). As we also know, lim(x^2/e^x) = 0. As we also also know, 12*0 = 0. So, here ya go.
As for millions or trillions of power, here's an inductive proof:
We'll prove that for any n, where n is a natural number, lim(x -> inf, x^n/e^x) = 0.
Basis step:
lim(x -> inf, x/e^x) = 0. This was proven in the video.
Inductive step:
Now, let's try to evaluate lim(x -> inf, x^n/e^x). If we plug in infinity, we get inf/inf. Therefore, we can use L'Hopital's rule.
The derivative of x^n = n * x^(n-1). So, lim(x -> inf, x^n/e^x) = lim(x -> inf, n * x^(n-1)/e^x).
As per limit laws, lim(x -> inf, n * x^(n-1)/e^x) = n * lim(x -> inf, x^(n-1)/e^x).
We take for granted that lim(x -> inf, x^(n-1)/e^x) = 0.
n * lim(x -> inf, x^(n-1)/e^x) = n * 0 = 0
Since we know that lim(x -> inf, x^n/e^x) = 0 for n = 1, and we know that if lim(x -> inf, x^(n-1)/e^x) = 0 then lim(x -> inf, x^n/e^x) = 0, we know that lim(x -> inf, x^n/e^x) = 0 for all natural numbers.
isn't it L'hospital rule?
wouldn't (e^x/x^2) give another infinite/infinite?
@@DrTrefor Thank you I understand! Great videos by the way.