How to find #lim(ln(2x)/(2+x)) x-> # infinity?

2 Answers
Sep 3, 2017

The limit equals #0#

Explanation:

Since we are of the form #oo/oo#, we can use L'Hospitals rule.

#L = lim_(x->oo) (2/(2x))/(1)#

#L = lim_(x->oo) 1/x#

This is now a recognizable and commonly seen limit.

#L = 0#

Hopefully this helps!

Sep 4, 2017

#lim_(xrarroo)ln(2x)/(2+x)=0#

Explanation:

Logarithmic functions grow slower than polynomial functions. Polynomial functions grow slower than exponential functions.

Since #ln(2x)# is logarithmic and #2+x# is a polynomial (it only has a degree of #1#, but it's still a polynomial), the polynomial in the denominator will grow faster than the logarithmic function in the numerator.

Thus, the denominator will outpace the numerator and as #x# goes to infinity, the limit approaches #0#.

Note what would happen if the fraction were inverted:

#lim_(xrarroo)ln(2x)/(2+x)=0#

#lim_(xrarroo)(2+x)/ln(2x)=oo#