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I want to be clear with limits of even index root function like this one: $f(x)=\sqrt{x-1}$. The domain of this function is $[1,\infty)$.

If I'm looking for one-sided limits:

$$\lim _{x\to 1^{-}}\left(\sqrt{x-1}\right)=DNE$$ doesn't exist because the function is not defined at the left of $x=1$.

$$\lim _{x\to 1^{+}}\left(\sqrt{x-1}\right)=0$$ does exist because the function is defined at the right of $x=1$

Then,

$$\lim _{x\to 1}\left(\sqrt{x-1}\right)=DNE$$ doesn't exist because the one sided limits are not equal.

Is my analysis right?

bdvg2302
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1 Answers1

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Given a real function $f:D\to\mathbb{R}$ where $D\subseteq\mathbb{R}$ we say that, for a point $x_0\in D$, $\lim\limits_{x\to x_0}f(x_0)=L$ if $$\forall\epsilon>0\ \exists\delta>0\ \forall x\in D\cap(x_0-\delta,x_0+\delta)\setminus\{x_0\}:|f(x)-L|<\epsilon$$ that is if, zooming in sufficiently on the domain, we can make the image of every value around $x_0$ get as close as desired to the real value $L$. On the other hand, the sided limits $\lim\limits_{x\to x_0^-}f(x_0)=L$ and $\lim\limits_{x\to x_0^+}f(x_0)=L$ are respectively defined as $$\forall\epsilon>0\ \exists\delta>0\ \forall x\in D\cap(x_0-\delta,x_0):|f(x)-L|<\epsilon$$ $$\forall\epsilon>0\ \exists\delta>0\ \forall x\in D\cap(x_0,x_0+\delta):|f(x)-L|<\epsilon$$ The thing is, in your case, since $D=[1,+\infty)$ and $x_0=1$, it turns out that the limits $\lim\limits_{x\to1^+}f(x)=\lim \limits_{x\to1}f(x)$ (and, in your case, they both exist) directly due to the fact that $$D\cap(x_0-\delta,x_0+\delta)\setminus\{x_0\}=(1,1+\delta)= D\cap(x_0,x_0+\delta) $$ while the limit $\lim\limits_{x\to1^-}f(x)$ isn’t well determined since every point on the domain is $\geq x_0$ so $D\cap(x_0-\delta,x_0)=\emptyset$.

K. Makabre
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  • The definition in most calculus books requires that the function be defined on a punctured neighborhood of $x_0$; if that is not the case, the limit does not exist. While your definition is the more common one in more advanced courses, introducing it too early will cause confusion. I would suggest that you preface your answer with such a caveat: that this is a more general definition that may not be at work in a low-level course. – Arturo Magidin Feb 13 '23 at 14:57