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

##### 1 Answer

**Method 1: L'Hôpital's Rule**

The limit:

# lim_(t rarr oo) e^t/t #

is of an indeterminate form

# 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

**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.