Can the existence of a limit at x = 0 for sqrt(x) be debated?

[Caesar_Rahil]

can someone please convince me that lim x->0 sqrt(x) = 0
Who of you say it doesn't exist?

 

[statdad]

If you mean

<br /> \lim_{x \to 0^+} \sqrt{x}<br />

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

This

<br /> \lim_{x \to 0^-} \sqrt x<br />

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

<br /> \lim_{x \to 0} \sqrt x<br />

 

[tiny-tim]

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]

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]

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. )