What is the limit of #e^t / (3t^2)# as t approaches 0?

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Answer 1 · 2016-06-29T08:25:53

#oo#

#e^t / (3t^2) |_{t \to 0}#

as a practical matter, it's always useful to stuff on t = 0 and maybe #t = pm epsilon, epsilon > 0# to get an idea of whether it is 2 sided

#e^t / (3t^2) |_{t = 0} = 1/0# so not defined

and therefore #e^t / (3t^2) |_{t \to 0} = oo#

plus, because

#[e^epsilon ]_{epsilon to 0} = 1#

and

#[e^(-epsilon) ]_{epsilon to 0} = 1#

and because the denominator is in terms of #t^2#, so #epsilon^2 = (-epsilon)^2#, the limit should be the same approach from either side.