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How do you evaluate the limit #2t^2+8t+8# as t approaches #2#?

Answer 1

#lim_(t rarr 2) (2t^2+8t+8) = 32 #

If we define #f(t)=2t^2+8t+8#, then #f(t)# is continuous everywhere (ie it is well behaved and it has no jumps, discontinuities or places where the function is not defined).

Consequently

#lim_(t rarr a) f(t)= f(a) # for all values of #a#

hence

#lim_(t rarr 2) (2t^2+8t+8) = lim_(t rarr 2) f(t) # # " " = f(2) # # " " = 8+16+8 # # " " = 32 #

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Answer 2

Aurora Alvarado

#32#

The function is a polynomial and continuous thus the limit can be evaluated by substitution.

#lim_(t to2)(2t^2+8t+8)=8+16+8=32#

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Answer 3

Ezra Adams

To evaluate the limit of 2t^2 + 8t + 8 as t approaches 2, we substitute the value of 2 into the expression.

2(2)^2 + 8(2) + 8 = 2(4) + 16 + 8 = 8 + 16 + 8 = 32 + 8 = 40.

Therefore, the limit of 2t^2 + 8t + 8 as t approaches 2 is equal to 40.

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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.