Does $\lim_{x \to 5} \sqrt{x-5}$ exist, even though $5$ cannot be approached from the left? A basic limit question I don't quite understand... The question is as $$\lim_{x \to 5} \sqrt{x-5}$$ and solve for the limit.
When I plug in $5$ I get the answer zero, but when I tackle this question graphically the limit does not exist as nothing comes from the left of $5$.

Could anyone clear this up for me? Thanks.
It is a common mistake to forget an essential aspect of the definition of the limit: the constraint $|f(x)-L|&lt;\epsilon$ must hold $\forall x\color{red}{\in\text{dom}(f)\setminus\{c\}}:|x-c|&lt;\delta$.
I would say &quot;Yes&quot; but I prefer the answer &quot;It depends upon the definition/notation of limit we are referring to&quot;.
With reference to the simplified definition, often used in high school context, that is
$$\lim_{x \to x_0} f(x)=L \iff \lim_{x \to x_0^-} f(x)=\lim_{x \to x_0^+} f(x)=L$$
we should conclude that limit doesn't exists but, according to the more general definition of limit, excluding from the domain the points where the expression is not defined, we have that

$$\lim_{x \to 5} \sqrt{x-5}=:\lim_{x \to 5^+} \sqrt{x-5}=0$$

and we can conclude that the limit exists.
Refer also to the related

What is $\lim_{x \to 0}\frac{\sin(\frac 1x)}{\sin (\frac 1 x)}$ ? Does it exist?
A limit exists iff and only the left limit and the right limit exist and are equal to each other


Here is an example of &quot;simplified&quot; definition

(credits)