finding the limit using l'hopital's rule?

This takes the form infinity/infinity. If we apply L'Hopital's Rule, we get:

lim (x-->infinity) (2000x^1999)/e^x, by differentiating the numerator and denominator

= 2000 * lim (x-->infinity) x^1999/e^x.

If we apply L'Hopital's Rule again, we get:

2000 * lim (x-->infinity) (1999x^1998)/e^x

= 2000 * 1999 * lim (x-->infinity) x^1998/e^x.

As you can see, we can just keep applying the rule as long as there is an x term in the numerator.

The objective is to apply L'Hopital's Rule until the numerator becomes constant. This will require 2000 L'Hopital's Rule applications, which, obviously, applying it 2000 times is a bit ridiculous. However, if you note that the 2000th derivative of x^2000 is 2000 * 1999 * ... * 2 * 1 = 2000! and the 2000th derivative of e^x is just e^x, then:

lim (x-->infinity) x^2000/e^x

= 2000! * lim (x-->infinity) 1/e^x

= (2000!)(0)

= 0.

I hope this helps!

If apply L'Hopital's rule 2001 times you get 0/e^x = 0

0.

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