Using L'Hopital's Rule to show that exponentials dominate polynomials

[@bp56789]

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.

[@empru7029]

THIS IS SO HELPFUL TY!! you’re wonderful at explaining things

[@iqbalmaruf4944]

So good, sir. Request for Cauchy's Mean Value Theorem for L'Hopital's Rule.

[@rvmishra9881]

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.?

[@toomanycharacter]

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.

[@niallomurchu285]

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.

[@weisanpang7173]

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.

[@funnyconceptsi3919]

Interesting and fun Love from punjab india

[@rajdeeppaul333]

isn't it L'hospital rule?

[@FruitloopLeviathan]

wouldn't (e^x/x^2) give another infinite/infinite?

[@FruitloopLeviathan]

@@DrTrefor Thank you I understand! Great videos by the way.