Expert: Ahmed Salami - 12/9/2009

Question
QUESTION: is sqrt(|x|) continuous at x=0?
what I think
f(0)=sqrt|0|=0

g(x)=|x|
h(x)=sqrt(x)
f(x)=h(g(x)) =
sqrt(g(x)) is continuous if g(x) >or=0 [right?]
g(x)= |x|
= x if x >or=0
=-x if x< 0
g(0)=|0|=0
g(x) =0
h(x) is continuous when x=0
f(x) is continuous when x=0
but if I do this
lim x->0+ sqrt(|x|), since x>0 |x|=x =sqrt(x)
limx ->0- |x|=-x =sqrt(-x) /= sqrt (x)
f(x) isnt continuous when x =0
please help I know it's a little messy.


ANSWER: Hi Hamad,
Why did you conclude that f(x) isnt continuous when x = 0?
Clearly, √(0) = √(-0) = 0
and it is in fact continuous.
A function f(x) is continuous at x = a if the limit exists there and is equal to the function value at the point, i.e
lim x→a f(x) = f(a)
lim x→0 √|x| = 0
since lim x→0+ √|x| = lim x→0- √|x| = 0
f(0) = √|0| = 0
And so we have proven continuity at x = 0

Regards

---------- FOLLOW-UP ----------

QUESTION: because the |x| has to be defined
and as x ->0- |x| =-x
lim x→0- √-x  which isnt the same as lim x→0+ √x

Answer
Hi Hamad,
Of course its the same.
lim x→0- √-x = 0
lim x→0+ √x = 0
Am i missing something here? What is it exactly that you disagree with?
I'm attaching a graph of the function and you can clearly see that it is continuous.

Regards