Given to solve ,

`lim_(x->0^(+)) (e^x - (1+x)) / x^3`

as `x->0+` then the `(e^x- (1+x)) / x^3=0/0` form

so upon applying the L 'Hopital rule we get the solution as follows,

as for the general equation it is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Given to solve ,

`lim_(x->0^(+)) (e^x - (1+x)) / x^3`

as `x->0+` then the `(e^x- (1+x)) / x^3=0/0` form

so upon applying the L 'Hopital rule we get the solution as follows,

as for the general equation it is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by using the L'Hopital Rule we get the solution with the below form.

`lim_(x->a) (f'(x))/(g'(x))`

so, now evaluating

`lim_(x->0^(+)) (e^x - (1+x)) / x^3`

=`lim_(x->0^(+)) (e^x - (1+x))' / (x^3)'`

= `lim_(x->0^(+)) ((e^x - 1)) / ((3x^2))`

When `x->0+` we get `(e^x - 1) /(3x^2) = 0/0` form, so applying the l'Hopital's Rule again we get

= `lim_(x->0^(+)) ((e^x - 1)') / ((3x^2)')`

=` lim_(x->0^(+)) (e^x) / ((6x))`

so now plugging the vale of `x= 0` we get

= `lim_(x->0^(+)) (e^x) / ((6x))`

= `(e^0) / ((6(0)))`

= ` 1/0`

`= oo`