Defined or undefined limit?

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I'm having trouble finding what the value of $\lim_{x \to 1^+} \sqrt{1-x}$ is.

Is it zero, or is it undefined? I asked in another forum as well, and received different opinions.. I am very confused.

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2 Answers

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It is undefined.

The function $x\mapsto \sqrt{1-x}$ is only defined for $1-x>0$, so only for $x<1$.

In your case, the limit only takes undefined values, since it is a limit when $x$ is decreasing to $1$.

The limit $$\lim_{x\to 1^-}\sqrt{1-x}$$ on the other hand does exist and is equal to $0$.

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In fact, the square root $\sqrt{1-x}$ is meaningless for all $x > 1$. So, in particular, the limit $\lim_{x \to 1+}\sqrt{1-x}$ does not exist at all.

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