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

1 Answer
May 22, 2016

through l'Hôpital's rule, find #8/3#

Explanation:

We have:

#lim_(xrarr6)(x-2sqrt(x+3))/(x-3sqrt(x-2))#

When #6# is plugged in, this results in the indeterminate form #0/0#, thus l'Hôpital's rule can be applied. Taking the derivative of the numerator and denominator, we see that

#=lim_(xrarr6)(1-1/sqrt(x+3))/(1-3/(2sqrt(x-2)))#

Evaluating the limit, this becomes

#=(1-1/sqrt(6+3))/(1-3/(2sqrt(6-2)))=(1-1/3)/(1-3/4)=(2/3)/(1/4)=8/3#