# Answer to Question #24325 in Calculus for sam

Question #24325

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

Expert's answer

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

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

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