Answer to Question #24325 in Calculus for sam

Question

i want to evaluate the limit as x goes to 2 for ((radical 6-x)-2) divided by ((radical 3-x)-1) please

Answer 1

Lim[ (Sqrt[6-x]-2)/(Sqrt[3-x]-1), x->2]
If x->, then
2-x<<1
For example, 2-x=a, a->0
Therefore:
Lim[ (Sqrt[6-x]-2)/(Sqrt[3-x]-1), x->2] = Lim[ (Sqrt[4-a]-2)/c(Sqrt[1-a]-1), a->0 ]
Taylor series:
Sqrt[4-a] = 2Sqrt[1-a/4] ~ 2*(1-(a/4)/2) = 2 - a/4
Sqrt[1-a] ~ 1 - a/2
(Sqrt[4-a]-2)/c(Sqrt[1-a]-1) = (2 - a/4 - 2)/(1-a/2 -1) = (a/4)/(a/2) = 1/2
Answer: Lim = 1/2