Question

How do you find the limit of `e^(x-x^2)` as x approaches infinity?

Answer 1

We know `x` approches `+prop`
Thus `x²` also approches `+prop`, but faster
Thus `x-x²` approches `-prop`
And we know that when `y` approches `-prop`, `e^y` approches `0`
Thus `lim(e^(x-x²))=0`

Answer 2

`y = lim_(x->oo) e^(x-x^2)`

`ln y = lim_(x->oo) (x-x^2) = -oo`

graph{x-x^2 [-5.67, 14.33, -9.08, 0.92]}

`=> y = lim_(x->oo) e^(-oo) = 1/(oo) = 0`