Question #8536a

1 Answer
Jim H
Mar 25, 2017

The limit is #0#.

Explanation:

#lim_(xrarr-oo) x^2e^(8x)# has indeterminate initial form #oo * 0#

Rewrite it to use l'Hospital's Rule

#lim_(xrarr-oo) x^2/e^(-8x)# Has intial form #oo/oo# so we can apply l'Hospital.

# (d/dx(x^2))/(d/dx(e^(-8x))) = (2x)/(-8e^(-8x))# which has form #-oo/-oo# so use llHospital again.

#(d/dx(2x))/(d/dx(-8e^(-8x))) = 2/(64e^(-8x)) = 1/32e^(8x)# which goes to #0# as #xrarr-oo#

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