What is the limit of #ln(x) / sqrtx # as x approaches #oo#?

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Answer 1 · 2017-08-22T13:32:42

The limit is #0#.

Short method

#lnx# grows more slowly than any positive power of #x#.

That is, for any positive #p#, #lim_(xrarroo)lnx/x^p = 0#

l'Hospital's rule

#lim_(xrarroo)lnx/sqrtx# has indeterminate initial form #0/0#, so we can use l'Hospital.

#lim_(xrarroo)(1/x)/(1/(2sqrtx)) = lim_(xrarroo)(2sqrtx)/x = lim_(xrarroo) 2/sqrtx = 0#

So #lim_(xrarroo)lnx/sqrtx = 0#