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

Question

2611 views around the world

You can reuse this answer
Creative Commons License

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#