How do you find the limit #lim 10^(2t-t^2# as #t->oo#?

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Answer 1 · 2017-05-21T15:05:31

#lim_(t->oo)# #10^(2t-t^2)=0#

As #t->oo#, #2t-t^2->-oo#

#therefore lim_(t->oo)# #10^(2t-t^2)=10^-oo#

#10^(-oo)=1/10^oo=0#

Answer 2 · 2017-05-21T15:10:40

#lim_(t->oo) 10^(2t-t^2) = 0#

#lim_(t->oo) 10^(2t-t^2)#

#= lim_(t->oo) 1/(10^(t^2-2t))#

By direct subtstitution
#= 1/10^oo#

# = 0#