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Answer 1 · 2016-12-21T17:19:32

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

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 $oo$ .