Q: If the function is continuous at some point, then is it necessary that limit should exist at that point? If the function is continuous at some point, then is it necessary that limit should  exists at that point?
Like in case of $\sqrt{x}$, its continuous at $x=0$, but limit doesn't exist as left hand limit is not defined.
But it appears meaningless to say that function is continuous at some point, but limit doesn't exist at that point.
Any inputs?

A: Note that $f(x)=\sqrt{x}$ is defined on $[0,\infty)$, to stay accord with the continuity-limit property, we talk about 
\begin{align*}
\lim_{x\rightarrow 0,x\in[0,\infty)}f(x),
\end{align*}
which is essentially the right-sided limit of $f$ at $0$. The running variable must at least stay in the domain of the function, or else it just doesn't make sense.

A: If you're familiar with $\varepsilon-\delta$ arguments:
Recall that we say a function $f$ is continuous at a point $x_0$ if for all $\varepsilon>0$, there exists $\delta>0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\varepsilon$. For the function $f$ given by $f(x)=\sqrt x$, no such $\delta$ can exist, since for any negative $x$ a distance less than $\delta$ from $x_0$, $f(x)$ is undefined. So $f$ is not continuous at $0$.
We can however say that $f$ is right-continuous at $0$, or equivalently, for all $\varepsilon>0$ there exists $\delta>0$ such that if $0\leq x-x_0<\delta$, then $|f(x)-f(x_0)|<\varepsilon$. Then $\sqrt x$ satisfies the definition of right-continuity at the origin. Recalling the definition of the right-sided limit, you can convince yourself that $f$ is right-continuous at $x_0$ if and only if the limit as $x\to x_0^+$ exists.
