How do you evaluate the limit #lim e^t/t# as #t-oo#?

Question

How do you evaluate the limit #lim e^t/t# as #t->oo#?

Luke Alvarez · Accepted Answer

Method 1: L'Hôpital's Rule The limit: # lim_(t rarr oo) e^t/t # is of an indeterminate form #oo/oo#, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exits then: # lim_(x rarr a) f(x)/g(x)  = lim_(x rarr a) (f'(x))/(g'(x)) # And so applying L'Hôpital's rule we get: # lim_(t rarr oo) e^t/t = lim_(t rarr oo) (d/dt e^t ) / ( d/dt t )#  # "           "= lim_(t rarr oo) (e^t ) / ( 1 )# 
# "           "= oo \ \ \ \# as #e^t# is unbounded. Method 2: Graphically graph{(e^x)/x [-26.97, 23.66, -2.97, 22.35]}  Nothing more to say; clearly #y=e^x/x# is unbounded Method 3: Taylor Series # e^t/t = 1/t \ e^t #
# \ \ \ \ = 1/t {1+t+t^2/(2!) + t^3/(3!) + t^4/(4!) + ... }#
# \ \ \ \ = 1/t+1+t/(2!) + t^2/(3!) + t^3/(4!) + ... # So then: # lim_(t rarr oo) e^t/t = lim_(t rarr oo) {1/t+1+t/(2!) + t^2/(3!) + t^3/(4!) + ... }#
# "           "= lim_(t rarr oo) {1/t+1 +O(t) }#
# "           "= lim_(t rarr oo) {1/t+1} +  lim_(t rarr oo){O(t) }# Although the first limit is finite, the second is unbounded.

Ezra Adams · Answer

undefined The limit of e^t/t as t approaches infinity is equal to zero.

How do you evaluate the limit #lim e^t/t# as #t-&gt;oo#?

How do you evaluate the limit #lim e^t/t# as #t->oo#?