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

2 Answers
Tessalsifi
Jun 2, 2015

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#

Truong-Son N.
Jun 11, 2015

#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#

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