It is given `f(x)=e^x+x^3-x^2+x`

we need to find `lim_(x->oo)((f^(-1)(x))/ln(x))`

`` Let we assume `f^(-1)=g` ,we know

`g(f(x))=x` (i) differentiate w.r.t x ,we have

`g'(f(x))f'(x)=1`

`g'(f(x))=1/(f'(x))`

Furthe let us assume that

`lim_(y->oo)g(y)=oo` where y=f(x) i.e ,for all y in its domain of definition ,in particular if y=x.

we assumming that `lim_(x->oo)f(x)=oo , means, y->oo` There are many function where this assumtion does not hold ,for example we can say about `tan ,tan^(-1)`

`lim_(x->oo,y->oo)(g(y)/ln(x))=oo/oo` ,apply L'Hospitals rule

`=lim_(x->oo)(1/(f'(x)))/((lnx)')`

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

Again apply L'Hospitals rule.

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

`` `=lim_(x->oo)(1/(e^x+6x-2))`

`=0`

`lim_(x->oo)((f^(-1)(x))/ln(x))=0`

``

`` `(f(x))'` denotes derivative of f(x) w.r.t. x