Given to solve,

`lim_(x->oo) (x^2 +4x+7)/(x-6)`

as `x->oo` then the ` (x^2 +4x+7)/(x-6) =oo/oo` is of indeterminate form

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

If `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->oo) (x^2 +4x+7)/(x-6)`

= `lim_(x->oo) ((x^2 +4x+7)')/((x-6)')`

= `lim_(x->oo) (2x+4)/(1)`

by plugging the value `x=oo` , we get

=` 2(oo)+4`

= `oo`

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