Question

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

Answer 1

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`