## question of basic calculus

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

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 Recognitions: Homework Help If you mean $$\lim_{x \to 0^+} \sqrt{x}$$ (limit as x approaches 0 through the positives), then the limit is zero. Notice that this is a one-sided limit. This $$\lim_{x \to 0^-} \sqrt x$$ 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 $$\lim_{x \to 0} \sqrt x$$

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 Quote by Caesar_Rahil 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?

## 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?

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