Question

How do you find the limit of `(1-[6/x])^x` as x approaches infinity using l'hospital's rule?

Answer 1

`lim_(x->oo)(1-6/x)^x= 1/e^6`

Explanation:

`lim_(x->oo)(1-6/x)^x = lim_(x->oo)e^ln((1-6/x)^x)`

`=lim_(x->oo)e^(xln(1-6/x))`

`=e^(lim_(x->oo)xln(1-6/x))" (*)"`

(Note that the above step holds as `e^x` is a continuous function)

`lim_(x->oo)xln(1-6/x) = lim_(x->0)ln(1-6x)/x`

As the above is a `0/0` indeterminate form , we can apply L'hopital's rule.

`lim_(x->0)ln(1-6x)/x = lim_(x->0)(d/dxln(1-6x))/(d/dxx)`

`=lim_(x->0)(-6/(1-6x))/1`

`=-6`

Substituting this back in for `lim_(x->oo)xln(1-6/x)` in `"(*)"` gives us

`lim_(x->oo)(1-6/x)^x = e^(-6) = 1/e^6`