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

Answer 1

Christopher Allers

#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#

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Answer 2

Christopher Allers

#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#

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Answer 3

Isabella Ackerman

To find the limit of the function 10^(2t-t^2) as t approaches infinity, we can analyze the behavior of the function as t becomes larger and larger.

As t approaches infinity, the term t^2 grows much faster than 2t. Therefore, the exponent 2t-t^2 becomes dominated by the negative t^2 term.

Since any positive number raised to a negative power approaches zero as the exponent becomes larger, we can conclude that the limit of 10^(2t-t^2) as t approaches infinity is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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