Question

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

Answer 1

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.