the limit is -1
Apply l'Hôpital's Rule twice
Find the limit. (If the limit is infinite, enter '∞' or '-∞', as appropriate. If the limit does not otherwise exist, enter DNE.)
lim t-inf (sqrt(t)+t^2)/(9t-t^2)
2 answers
Divide numerator and denominator by t^2
(sqrt(t)/t^2+t^2/t^2)/(9t/t^2-t^2/t^2) = (1/t^(3/2)+1)/(9/t -1)
lim 1/t^(3/2)=0
t->∞
lim 9/t=0
t->∞
also
lim 1/t^(3/2)=0
t-> -∞
lim 9/t=0
t-> -∞
So
lim (1/t^(3/2)+1)/(9/t -1) =
t-> -∞
(0+1)/(0-1) = 1/ - 1 = - 1
lim (1/t^(3/2)+1)/(9/t -1) =
t-> ∞
(0+1)/(0-1) = 1/ - 1 = - 1
(sqrt(t)/t^2+t^2/t^2)/(9t/t^2-t^2/t^2) = (1/t^(3/2)+1)/(9/t -1)
lim 1/t^(3/2)=0
t->∞
lim 9/t=0
t->∞
also
lim 1/t^(3/2)=0
t-> -∞
lim 9/t=0
t-> -∞
So
lim (1/t^(3/2)+1)/(9/t -1) =
t-> -∞
(0+1)/(0-1) = 1/ - 1 = - 1
lim (1/t^(3/2)+1)/(9/t -1) =
t-> ∞
(0+1)/(0-1) = 1/ - 1 = - 1
0 answers