# calculate limit after transform `y=f(x)``limit (f^(-1)(x))/(ln x)` x go to positive infinite `f(x) = e^x + x^3 - x^2 + x`

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

Since the problem requests to perform the transformation `y = f(x)` , you may evaluate the limit, such that:

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

Since `y = f(x)` yields:

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

SinceĀ `(f^(-1)(f(x))) = x` and substituting `e^x + x^3 - x^2 + x` for `f(x)` yields:

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

Substituting `oo` for x in limit yields:

`lim_(y->oo) x/(ln(e^x + x^3 - x^2 + x)) = oo/oo`

You may use l'Hospital's theorem such that:

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

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

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

Factoring out `e^x` yields:

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

Reducing duplicate factors yields:

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

Using the information that `lim_(x->oo) x^n/e^x = 0` yields:

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

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

Hence, evaluating the limit, under the given conditions, yields `lim_(x->oo) (f^(-1)(x))/(ln x) = 1.`