How do you evaluate # (3+10x)/(6x-10)# as x approaches infinity?

1 Answer
Steve M
Oct 10, 2016

#lim_(xrarroo)(3+10x)/(6x-10)=5/3#

Explanation:

#lim_(xrarroo)(3+10x)/(6x-10)=lim_(xrarroo)((3+10x)/x)/((6x-10)/x)#
# =lim_(xrarroo)(3/x+10)/(6-10/x)#

We now observe that as #x->oo# then #3/x->0 and 10/x->0#

Hence, #lim_(xrarroo)(3+10x)/(6x-10)=10/6=5/3#

This can be visualized by looking at the graph for large #x#
graph{(3+10x)/(6x-10) [-10, 60, -5, 5]}

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