First we need to show that

`lim_(x->oo) (lnx)^n/x = 0` for all positive n, since this is oo/oo we can use L'Hopital's rule to get

`lim_(x->oo) (n(lnx)^(n-1)(1/x))/1 = lim_(x->oo) (n(lnx)^(n-1))/x`

If you keep taking derivatives you will eventually get that this limit

`lim_(x->oo) (lnx)^n/x = lim_(x->oo) (n! lnx)/x`

Taking another L'Hopital, we get

` lim_(x->oo) (lnx)^n/x = lim_(x->oo) (n!)/x`

and this limit is equal to zero so

`lim_(x->oo) (lnx)^n/x = 0`

It is also obvious that if `lim_(x->oo) (lnx)^n/x = 0` then `lim_(x->oo) (lnx)^n/x^m=0` for all positive m and n.

Going to our original limit we can divide everything by `x^3` to get

`lim_(x->oo) (9x^3.01 + 25x)/(3x^3+10(lnx)^100) = lim_(x->oo) (9x^0.01+25/x^2)/(3+10(lnx)^100/x^3)` Now as `x->oo` every item except for `9x^0.01` and `3` go to zero as `x->oo` so we get`lim_(x->oo) (9x^3.01+25x)/(3x^3+10(lnx)^100) = lim_(x->oo) (9x^0.01)/3` and this limit is oo.So `lim_(x->oo) (9x^3.01+25x)/(3x^3+10(lnx)^100) = oo`## We’ll help your grades soar

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