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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#

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Answer 2

William Abdalla

To find the limit of (sqrt(6-x)-2)/(sqrt(3-x) -1) as x approaches 2, we can use algebraic manipulation. First, we simplify the expression by multiplying both the numerator and denominator by the conjugate of the denominator, which is sqrt(3-x) + 1. This will help eliminate the square roots in the expression. After simplifying, we get (6-x-4)/(3-x-1). Further simplifying, we have (2-x)/(2-x). Now, we can cancel out the common factor of (2-x) in the numerator and denominator. The expression simplifies to 1. Therefore, the limit of (sqrt(6-x)-2)/(sqrt(3-x) -1) as x approaches 2 is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.