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`lim_(x->0^(+)) (e^x- (1+x)) / x^3` Evaluate the limit, using L’Hôpital’s Rule if necessary.

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`

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