Question

Question #af52c

Answer 1

`lim_(xrarroo)e^(3x)/(5x^2) = oo`. Because `e^x` grows faster that any positive power of `x`.

Explanation:

The quick explanation is that `lim_(xrarroo)e^x/x^p = oo` for any positive `p`. (Indeed, for any real `p`.)

My longer explanation uses l'Hospitals's Rule

`lim_(xrarroo)e^(3x)/(5x^2)` has indeterminate initial form `oo/oo`.

Apply l'Hospital and consider

`lim_(xrarroo)(3e^(3x))/(10x)` which also has indeterminate initial form `oo/oo`.

Apply l'Hospital again to get

`lim_(xrarroo)(9e^(3x))/(10)` which is `oo`.

Therefore the original limit is `oo`.