you have to put expression as the ratio of two functions firstsacwchiri said:Im trying to solve this problem using l'hopital but amm not sure how to do it
lim
X->infinite x^3 * e^(-x^2)
soo this infinite * e^-infinite... but from there I am not sure if you can use it to solve this...
sacwchiri said:yeah i thought about that but amm not sure how much is e^-infinite
That sounds like you are saying "how infinite is it"! What you mean is "what number is e^-infinite". (Strictly speaking that is also meaningless- you cannot evaluate a function of real numbers "at infinity", you can only take limits at infinity.) You know, I hope, that e^x increases without bound (i.e. "goes to infinity") as x goes to infinity. You should also know that "e^(-x2) MEANS 1/e^(x2). If A goes to infinity, what does 1/A go to?sacwchiri said:yeah i thought about that but amm not sure how much is e^-infinite
L'Hôpital's Rule is a mathematical theorem used to solve indeterminate forms in calculus, particularly in finding limits of functions. It states that for certain types of limits, the limit of the ratio of two functions is equal to the limit of their derivatives.
L'Hôpital's Rule is used when evaluating limits that result in indeterminate forms such as 0/0 or ∞/∞. It can also be used when evaluating limits of the form 0 x ∞.
The steps for using L'Hôpital's Rule to solve an infinite limit are as follows:
No, L'Hôpital's Rule can only be applied to certain types of infinite limits, specifically those that result in indeterminate forms such as 0/0 or ∞/∞. It cannot be used for limits that do not result in these forms.
Yes, L'Hôpital's Rule can only be used for functions that are differentiable in the given interval. Additionally, it can only be used for limits that result in indeterminate forms and may not always provide a definitive answer.