How do you find the Limit of #ln(x) / sqrtx # as x approaches infinity?

Question

4029 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-02T18:33:16

#ln(x)# grows more slowly than any positive power of #x# as #x->oo#.

Hence the limit is #0#

Quick answer

Note that the exponential function #e^x# grows faster than any polynomial, in particular #x^2#

As a result the inverse function #ln x# grows more slowly than the inverse function #sqrt(x)#

Here's #e^x# and #x^2# ...

graph{(y-e^x)(y-x^2) = 0 [-39.96, 40.04, -164.4, 235.6]}

Here's #ln x# and #x^(1/2)# ...

graph{(y - ln x)(y - x^(1/2)) = 0 [-96.4, 100.36, -4.56, 5.44]}

Hence:

#lim_(x->oo) (ln x)/(sqrt(x)) = 0#