Checking Continuity of functions

Question

Consider a function $\sqrt {x-1}$+$\sqrt{1-x}$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions

Answer 1

What is your definition for $l=\lim\limits_{x\to 1}f(x)$ if it exists?

If this is it:

$$\forall\varepsilon>0,\exists\delta>0,\forall x\in D_f,|x-1|<\delta\implies|f(x)-f(1)|<\varepsilon$$

then take ANY $\varepsilon>0$ and ANY $\delta>0$. $D_f=\{1\}$ so $|x-1|=0$ for any $x\in D_f$ and thus $f(x)=f(1)$.

Answer 2

Continuity is a topological concept.

If $\langle X,\tau_X\rangle$ and $\langle Y,\tau_Y\rangle$ are topological spaces and $f:X\to Y$ is a function then $f$ is by definition continuous if all preimages of elements $\tau_Y\subseteq\wp(Y)$ under $f$ are elements of $\tau_X\subseteq\wp(X)$.

So that is the criterion that should be handled here.

Observe that limits do not play a part in it.

In this situation $X=\{1\}$ and then automatically we must have $\tau_X=\wp(X)$ (there are no other topologies on $\{1\}$).

Then it is immediate that any function that has $\{1\}$ as domain is continuous, because preimages under that function will of course be elements of $\wp(\{1\})=\{\varnothing,\{1\}\}$.

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It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.

To get more view on this you could take a look at this answer.

Answer 3

If you're using the definition of $\lim_{x \to a} f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $\varepsilon$/$\delta$ definition of being continuous at the point $a \in D_f$ if and only if

(1) $a$ is an accumulation point of the domain $D_f$, and $f(x) \to f(a)$ as $x \to a$

or

(2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $\lim_{x \to a} f(x)$).

In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.