Question

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

Answer 1

`= e^(ab)`

Explanation:

`lim_(x to oo) [1+(a/x)]^(bx)`

first you need to force it into an indeterminate form, and there's little to lose in trying to but the stuff in parentheses apart so we do this next step

`lim_(x to oo) exp ( ln [1+(a/x)]^(bx) )`

we can lift the `e` out as it is a continuous function
`exp ( lim_(x to oo) bx ln [1+(a/x)] )`

we also notice that `lim_(x to oo) ln [1+(a/x)] = ln 1 = 0`

so the indeterminate form is

`exp ( lim_(x to oo) (ln [1+(a/x)] )/(1/(bx)))` and we can apply L'Hopital

`= exp ( lim_(x to oo) (-a/x^2 1/[1+(a/x)] )/(-1/(bx^2)))`

`= exp ( lim_(x to oo) ( a b)/(1+(a/x)))`

`= e^(ab)`