How do you find the limit of #lnx/(sqrtx+lnx)# as #x->oo#?

1 Answer
mason m
Nov 6, 2016

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

Explanation:

#L=lim_(xrarroo)lnx/(sqrtx+lnx)#

This is an indeterminate form, in the form of #oo/oo#. Due to that, we can use L'Hopital's rule by differentiating the numerator and denominator individually.

#L=lim_(xrarroo)(d/dxlnx)/(d/dx(sqrtx+lnx))=lim_(xrarroo)(1/x)/(1/(2sqrtx)+1/x)#

Multiply through by #x/x#:

#L=lim_(xrarroo)(1/x*x)/((1/(2sqrtx)+1/x)*x)=lim_(xrarroo)1/(sqrtx/2+1)#

As we examine this, we see the denominator approaches infinity as the numerator stays unchanged, so the overall fraction will approach #0#.

#L=0#

Another way of approaching this is that both the numerator and denominator of the fraction contain #lnx#, so those will increase at the same rate. However, the denominator is also increasing with #sqrtx#, so the denominator will be growing faster than the numerator thanks to the additional #sqrtx#. Thus, the limit will be #0# because the denominator will constantly be outgrowing the numerator.

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