Question

What is `lim_(x->oo) e^x/(x-1)` ?

Answer 1

`oo`

Explanation:

`lim_(x->oo)e^x/(x-1)`

If we evaluate this by direct substitution we get

`e^oo/(oo-1)=oo/oo`

This answer is not helpful. Since `oo` is not a finite value, we don't know if this is equal to `1`, `9000008876`, `1/55`, etc. This is an indeterminate form.

L'Hospital's rule provides a solution for us. L'Hospital's rule states that:

`lim_(x->a)f(x)/g(x)=lim_(x->a)(f'(x))/(g'(x))` for indeterminate forms `oo/oo` and `0/0`

In other words, if and only if a limit ends up as `oo/oo` or `0/0`, you may take the derivative of the top and the derivative of the bottom and evaluate the limit of that. Please note that this is not the quotient rule. You are to take the derivative of the top and the bottom separately.

Applying the rule,

`lim_(x->oo)e^x/(x-1)=^Hlim_(x->oo)e^x/1=oo/1=oo`

Answer 2

`lim_(x->oo) e^x/(x-1) = oo`

Explanation:

The simple answer is that `e^x` grows faster than any polynomial as `x->oo`.

Hence:

`lim_(x->oo) e^x/(x-1) = oo`

A little more formally:

`e^x = sum_(k=0)^oo x^k/(k!) > x^2/2`

Hence:

`lim_(x->oo) e^x/(x-1) >= lim_(x->oo) (x^2/2)/(x-1) >= lim_(x->oo) 1/2(x^2/x) = lim_(x->oo) x/2 = oo`