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

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Answer 1 · 2016-11-03T14:13:19

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#

Answer 2 · 2016-11-03T14:16:41

#5/2#

The limit can be evaluated by direct substitution.

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