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$$\lim_{x \to 0} \sqrt{x}$$

I am not sure whether this limit approches to 1 or does not exist. According to Thomas Finny it does not exist.

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4 Answers 4

$\lim_{x \to 0+} \sqrt{x} = 0$. Never it will be 1

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The function $\sqrt{x}$ is continous on $[0,\infty)$, so it is simply $$\lim_{x\to0+}\sqrt{x}=\sqrt{0}=0.$$

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For a limit to exist it must be a real number$$\lim_{x \to 0^+} \sqrt{x} = 0$$ $$\lim_{x \to 0^-} = \text{DNE}$$

because $\sqrt{x} $ is not real for $x<0$

Since the left limit is not equal to the right limit , the total limit Does Not Exist.

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$\lim_{x\to 0} x$ does not exist, because $\lim_{x\to 0+} x=0$ but $x\neq 0$ for $x<0$. –  flonk Feb 28 '14 at 10:38
I don't see how that disqualifies my argument. For x < 0 $ \sqrt{x} $ is not real. –  neofoxmulder Feb 28 '14 at 10:39
$x$ is not $0$ for $x<0$, but it's left limit can be (and is) $0$. Likewise $\sqrt{x}$ is not real for $x<0$, but it's left limit can be (and is) $0$. –  flonk Feb 28 '14 at 10:42
My interpretation of the OP is that Thomas Finny is asking for the 2-sided limit , how can there be a 2-sided limit when you don't have 2 sides? :) –  neofoxmulder Feb 28 '14 at 11:10
@neofoxmulder, you are correct.. Anyways, this question is stupid. –  CobraCobra Feb 28 '14 at 11:51

Note that $\sqrt{x}$ (as a real-valued function) is not defined for $x<0$, so we can only talk about the limit coming from the right:

$$\lim_{x \to 0+} \sqrt{x}$$

Which exists and is zero.

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