How do you find the limit of $(2-x)/(sqrt(4-4x+x^2))$ as x approaches $2^+$ ?

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Answer 1 · 2016-04-20T13:04:22

Rewrite the expression using $sqrt(u^2) = absu$ and using the definition of the absolute value function.

First we write

$(2-x)/sqrt(4-4x+x^2) = (2-x)/sqrt((2-x)^2)$

$= (2-x)/abs(2-x)$ .

Now, note that

$abs(2-x) = { (2-x,"if",2-x >= 0),(-(2-x),"if",2-x < 0):}$

$= { (2-x,"if",x <= 2),(-(2-x),"if",x > 2):}$ .

Putting these together, we get

$(2-x)/sqrt(4-4x+x^2) = (2-x)/abs(2-x)$

$= { ((2-x)/(2-x) = 1,"if",x <= 2),((2-x)/(-(2-x))=-1,"if",x > 2):}$ .

As $xrarr2^+$ the function is constant $-1$

Therefore the limit is $-1$

How do you find the limit of (2-x)/(sqrt(4-4x+x^2))(2-x)/(sqrt(4-4x+x^2)) as x approaches 2^+2^+?