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Question of basic calculus 
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#1
Sep2309, 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??? 


#2
Sep2309, 01:31 PM

HW Helper
P: 1,371

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 onesided 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 onesided 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] 


#3
Sep2309, 01:34 PM

Sci Advisor
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Thanks
P: 26,148

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? 


#4
Sep2709, 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? 


#5
Sep2709, 10:00 AM

Sci Advisor
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Thanks
P: 26,148

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


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