Find the one sided limit of the expression $\sqrt{2 - x}$

Question

So I'm supposed to find the one sided limit of the following expression:

$\sqrt{2 - x}$ as $x$ approaches 2 from values greater than 2.

I can see that the square root will contain a value less than 0 and therefore the real limit does not exist. But does that mean no limit exists? I mean, can't I write the solution to this problem as $\lim_{x \rightarrow 2+} \sqrt{2 - x} = \lim_{b \rightarrow 0} = i b$ where $i = \sqrt{-1}$? Or does my solution equal 0? And therefore no one sided limit exist?

Answer 1

$$\lim_{x \rightarrow 2^+}\sqrt{2-x}$$

Know that

$$\sqrt{2-x} \geq0$$

$${2-x} \geq0$$

$$x\leq 2$$

$$\lim_{x \rightarrow 2^+}\sqrt{2-x}=DNE$$

Not for real number!

Let see for

$$\lim_{x \rightarrow 2^-}\sqrt{2-x}=\sqrt{2-2}$$ $$\lim_{x \rightarrow 2^-}\sqrt{2-x}=0 $$

Left hand limit does not equal to right hand limit

When we say that a limit exist it means

$$\lim_{x \rightarrow a}f(x)=\lim_{x \rightarrow a^+}f(x)=\lim_{x \rightarrow a^-}f(x)$$

Answer 2

If $\,\sqrt{z}\,$ is considered to be the principal value of the complex square root, then the limit exists in complex numbers $\,\lim_{z \to 2} \sqrt{2-z} = 0\,$. This follows because, if $\,z \to 2\,$, then $\,z-2 \to 0\,$, so $\,|z-2| \to 0\,$, therefore $\,|\sqrt{2-z}|=\sqrt{|2-z|} \to 0\,$, and also $\sqrt{2-z} \to 0$ since $|f| \to 0$ iff $f \to 0\,$.

Since the limit exists, it will hold for any complex path going to $\,2\,$, in particular for $x \to 2+$ on the real axis, so one could indeed write in that context that $\,\lim_{x \to 2+} \sqrt{2-x} = 0\,$. However, the interpretation (used above) under which this makes sense is not self-evident, so it must be very carefully stated if writing $\,\lim_{x \to 2+} \sqrt{2-x} = 0\,$, otherwise if taken out of context it just invites the kind of confusion often associated with the genre of $\,1=\sqrt{1^2}=\sqrt{(-1)^2}=-1\,$ "proofs".