How do you find the limit of #(sqrt(6-x)-2)/(sqrt(3-x) -1)# as x approaches 2?

1 Answer
Konstantinos Michailidis
Mar 12, 2016

Hence the limit #lim_(x->2) (sqrt(6-x)-2)/(sqrt(3-x) -1)=0/0# is an undefined form of #0/0# we can apply L'Hopital rule hence we get

#lim_(x->2) (sqrt(6-x)-2)/(sqrt(3-x) -1)= lim_(x->2) [(d(sqrt(6-x)-2))/dx]/[(sqrt(3-x) -1)/dx]= lim_(x->2) [-(1)/(2sqrt(6-x))]/[-(1)/(2sqrt(3-x)]]= lim_(x->2) [sqrt(3-x)/sqrt(6-x)]= 1/2#

Finally

#lim_(x->2) (sqrt(6-x)-2)/(sqrt(3-x) -1)=1/2#

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