When a function is defined at some point, does limit exist at that specific point? $$
f(x)=
\begin{cases}
x^2, & x<1\\
2.4, & x=1\\
x^2+1, &x>1
\end{cases}
$$
Does $\lim_{x\to 1}f(x)$ exist?
The values of one-sided limits are different for the function $f(x)$ at $x=1$, but at $x=1$ the function is, $f(x)=2.4$. My question is, does this function has limit at $x=1$? 
 A: Note that, for a limit to exist at a point,
$$\lim_{x \to c} f(x) = L \;\;\;\;\; \text{if and only if} \;\;\;\;\; \lim_{x \to c^+} f(x) = \lim_{x \to c^-} f(x) = L$$
Or, in words, the limit at a point exists and is equal to $L$, if and only if the limit from each side is also equal to $L$. (Note: it need not be true that the limit as $x \to c$ is equal to $f(c)$ for existence to occur.)
For example, consider the step function:
$$u(x) = \left\{ \begin{matrix}
0 & x \le 0 \\
1 & x > 0
\end{matrix} \right.$$
What is the limit as $x \to 0$? If you approach from the left, it's $0$, and if you approach from the right it's $1$. Therefore, the limit as $x \to 0$ for $u(x)$ doesn't exist, since you get a different limit of approach on each side, i.e.
$$\lim_{x \to 0} u(x) \; \text{doesn't exist, because}  \; 1 = \lim_{x \to 0^-} u(x) \ne \lim_{x \to 0^+} u(x) = 0$$
This can be seen graphically: see that "jump" at $x = 0$?

In your scenario, you're correct: the limit as $x \to 1$ from each side is different for your $f$. Therefore, you rightfully can conclude that the limit does not exist.

You can read more on the existence of limits on Brilliant.
