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Question of basic calculus

by Caesar_Rahil
Tags: basic, calculus
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Caesar_Rahil
#1
Sep23-09, 10:43 AM
P: 69
can someone please convince me that lim x->0 sqrt(x) = 0
Who of you say it doesn't exist???
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statdad
#2
Sep23-09, 01:31 PM
HW Helper
P: 1,361
If you mean

[tex]
\lim_{x \to 0^+} \sqrt{x}
[/tex]

(limit as x approaches 0 through the positives), then the limit is zero. Notice that this is a one-sided limit.

This

[tex]
\lim_{x \to 0^-} \sqrt x
[/tex]

means you are trying to approach 0 from the left - through negative numbers. I feel confident in saying this limit does not exist. Why might that be?

Finally, remember that the ordinary limit exists as a real number if, and only if, the two one-sided limits exist as real numbers and are equal. The comment and observation I made earlier combine to say (fill in the blank yourself) about

[tex]
\lim_{x \to 0} \sqrt x
[/tex]
tiny-tim
#3
Sep23-09, 01:34 PM
Sci Advisor
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Quote Quote by Caesar_Rahil View Post
can someone please convince me that lim x->0 sqrt(x) = 0
Who of you say it doesn't exist???
HI Caesar_Rahil!

lim x -> 0+ sqrt(x) = 0, but lim x->0- sqrt(x) depends on you define sqrt(x) for negative x

how are you defining it?

Do you define it as not existing, or do you define it as an imaginary number?

Caesar_Rahil
#4
Sep27-09, 09:40 AM
P: 69
Question of basic calculus

I'm confining myself to real valued functions
since left hand limit does not exit, that means sqrt(x) does not have a limit there
so it does not exist
is this interpretation correct?
tiny-tim
#5
Sep27-09, 10:00 AM
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Quote Quote by Caesar_Rahil View Post
I'm confining myself to real valued functions
since left hand limit does not exit, that means sqrt(x) does not have a limit there
so it does not exist
is this interpretation correct?
Yes if you define sqrt(x) for negative x as not existing, then automatically there cannot be a limit at x = 0.

(On the other hand, if you define sqrt as a function whose domain is the non-negative numbers only, then the limit does exist. )


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