# lim t>infinity (t^3-5t)^2/3t^5+2t-4show all working

### 1 Answer | Add Yours

You need to substitute oo for t in limit such that:

`lim_(t->oo) ((t^3-5t)^2)/(3t^5+2t-4) = oo/oo`

Since the indetermination is of type `oo/oo` , you may use l'Hospital's theorem such that:

`lim_(t->oo) ((t^3-5t)^2)/(3t^5+2t-4) = lim_(t->oo) (((t^3-5t)^2)')/((3t^5+2t-4)')`

`lim_(t->oo) ((t^3-5t)^2)/(3t^5+2t-4) = lim_(t->oo) (2(t^3-5t)(3t^2 - 5))/(15t^4 + 2)`

`lim_(t->oo) (2(t^3-5t)(3t^2 - 5))/(15t^4 + 2) = lim_(t->oo) (2(3t^5 - 20t^3 + 25t))/(15t^4 + 2)`

`lim_(t->oo) (2(3t^5 - 20t^3 + 25t))/(15t^4 + 2) = oo/oo`

You need to use again l'Hospital's theorem such that:

`lim_(t->oo) (2(15t^4 - 60t^2 + 25))/(60t^3) = oo/oo`

`lim_(t->oo) ((2(15t^4 - 60t^2 + 25))')/((60t^3)') = lim_(t->oo) (2(60t^3 - 120t))/(180t^2) = oo/oo`

`lim_(t->oo) (2(60t^3 - 120t))/(180t^2) = lim_(t->oo) ((2(60t^3 - 120t))')/((180t^2)')`

`lim_(t->oo) ((2(60t^3 - 120t))')/((180t^2)') = lim_(t->oo) (2(180t^2 - 120))/(360t) = oo/oo`

`lim_(t->oo) (2(180t^2 - 120))/(360t) = lim_(t->oo) ((2(180t^2 - 120))')/((360t)')`

`lim_(t->oo) (2(180t^2 - 120))/(360t) = lim_(t->oo) (2*360t)/360 = oo/360 = oo`

**Hence, evaluating the given limit using l'Hospital's theorem yields `lim_(t->oo) ((t^3-5t)^2)/(3t^5+2t-4) = oo` .**