Question

find limit as x approaches infinity of e^(x-x^2)
Do I use l'Hospitals rule for this and if so, how?

Answer 1

you could split it up as e^(x)*e^(-x^2) and apply lhospitals.
take derivative of e^(x)/e^(x^2)

Answer 2

Um you can't really use l'Hopital's rule here because you have a ratio of 2 exponentials:
\[\frac{e^x}{e^{x^2}}\] If you try differentitating the top and bottom, you'll end up in a never ending loop because the derivative of e^x is just e^x

Answer 3

No you can't use l'Hospital's rule. Realize by that the continuity of e^x, e^x at infinity is is just infinity, in this case it's negative infinity. Therefore you're evaluating to the negative infinity, which is zero.

Answer 4

but, you can do this:
\[\large \frac{\frac{e^e}{e^x}}{\frac{e^{x^2}}{e^x}}=\frac{1}{e^{x^2-x}}\]which goes to 0 as x goes to infinity