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?

  • $\begingroup$ I'm sorry, I did a test with my calculator and of course b does not approach 0. But my question still remains, is there no complex limit to this expression? $\endgroup$ – Aram Fatah Jul 9 '17 at 12:16
  • $\begingroup$ Please use MathJax. $\endgroup$ – José Carlos Santos Jul 9 '17 at 12:17
  • $\begingroup$ i would say this Limit doesn't exist $\endgroup$ – Dr. Sonnhard Graubner Jul 9 '17 at 12:23
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    $\begingroup$ If your square root function is defined for negative real numbers, then the limit exists and equals 0. If your square root function is not defined for negative real numbers, then the limit does not exist. $\endgroup$ – Wouter Jul 9 '17 at 12:29
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    $\begingroup$ You can write $$\sqrt{2-x}=i\cdot \sqrt{x-2}$$ So, if $x$ approaches $2$ from right, you can see that the function actually approaches $0$ in the sense you described. Your argument with the $b$ must be precised (how does $b$ depend on $x$ ?), but it should lead to the correct "limit". $\endgroup$ – Peter Jul 9 '17 at 12:34

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

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

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

  • $\begingroup$ Yes I know there exists no limit, but I was asking for the one-sided limit as a complex number? $\endgroup$ – Aram Fatah Jul 9 '17 at 12:33
  • $\begingroup$ It depends on your definition! If your criteria is in real number, then the limit does not exist for $x> 2$. $\endgroup$ – Crazy Jul 9 '17 at 12:35

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".


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