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How do you evaluate the limit #15/(t^2+5)# as t approaches #1#?

Answer 1

Christopher Agresta

Essentially, it is by substitution.

The long explanation is:

At #t# gets closer and closer to #1#, we know that #color(red)(t^2)# gets closer and closer to #color(red)1#

so #color(red)(t^2)+5# gets closer and closer to #color(red)(1)+5 = color(green)(6)#.

Therefore, #15/color(green)(t^2+5)# gets closer and closer to #15/color(green)(6)#

Finally, we can reduce the fraction #15/6 = (3*5)/(3*2) = 5/2#

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Answer 2

Elijah Abram

#5/2#

The limit can be evaluated by direct substitution.

#rArrlim_(t to1)15/(t^2+5)=15/6=5/2#

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Answer 3

Charles Anctil

To evaluate the limit of 15/(t^2+5) as t approaches 1, substitute 1 for t in the expression. The result is 15/(1^2+5) = 15/6 = 2.5.

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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.