# Calculus: Solve indeterminate question up

1. Nov 28, 2013

Does (infinity)^a * e^(-infinity), where a >1 equal to 0?

If so, why? If not, how to calculate it?

2. Nov 28, 2013

### Dick

You do these things by finding limits. 'infinity' by itself doesn't mean much. Do you mean limit x->infinity x^a*e^(-x)? Since you have an indeterminant limit exactly how you approach infinity in each instance of 'infinity' is pretty important.