No. You may have a function which has a discontinuity at x=a. Consider the function f(0) = 1 and f(x) = 0 if x is different from 0. It has 0 as both left and right limits and yet the function value at x=0 is one.

This is a different question and formulated with so many negations that I am not completely sure what question you actually intend to ask. Can you reformulate it?

Here you're asking about the notion of continuity of functions. A continuous function, by definition, has the property that its limit at a point is equal to its value at that point. But there are functions that are not continuous and so don't have this property. See the section Non-examples in the given link.

Sorry about that. It's a little difficult for me to express this abstract question in English.

The proposition is : [itex]x + 2 \neq 7[/itex] unless [itex] x = 5[/itex]
Is that proposition correct? I guess someone's answer is no because when x approaches 5, [itex]x + 2[/itex] would also be 7.

When x approaches 5, x + 2 approaches 7, but won't necessarily be equal to 7. When we talk about x approaching 5, what we're saying is that x is in some neighborhood around 5 (i.e., |x - 5| is "small"), so x + 2 will be in some neighborhood around 7 (i.e., |x + 2 - 7| is also "small"). x + 2 won't be equal to 7 if x ≠ 5.

No, for the function x+2, the statement is true. Even if it approaches 7 as x approaches 5, it is not 7 except for when x=5.

However, this is not always the case. Take the function f(x)=0 for all x. The limit when x goes to 0 is 0 and also f(0)=0. There are also examples where you will find that f takes the same values in very different points. That f(x) = a for only one x is a property called injective (also formulated as f(x) = f(y) implies x = y) and a linear function with non-zero slope has this property.

A better example would be the function f given by ##f(x) = \frac{x^2 + 3x + 2}{x + 1}##. We can talk about this limit:
##\lim_{x \to -1} f(x)##, which turns out to be 1, even though f is not defined at x = -1.

Mark44, thank you very much! You help me a lot. Now here's my next relevant question.
Actually, my question has something to do the following physics question:

In the uniform circular motion, is the instantaneous centripetal acceleration perpendicular to instantaneous velocity?

If we differentiate the velocity and solve the inner product [itex]\vec{a}\cdot \vec{v}[/itex], then we'll get zero.
However, if we imagine the [itex]\Delta \vec{v}[/itex] then we would think [itex]\vec{v}[/itex] will never be perpendicular to [itex]\Delta \vec{v}[/itex].
Someone would say we have to reconcile these two facts.

And I think the reason why they think those two argument are contradictory is that
they mistaken the direction of [itex]\vec{a}_{c}=\displaystyle \lim_{\Delta t\rightarrow 0}\frac{\Delta \vec{v}}{\Delta t}[/itex] as the direction of [itex]\Delta \vec{v}[/itex].

Just like what you told me, the direction of [itex]\vec{a}_{c}[/itex] is what [itex]\frac{\Delta \vec{v}}{\Delta t}[/itex] approaches as [itex]\Delta t[/itex] approaches [itex]0[/itex].
Therefore, the direction of [itex]\vec{a}_{c}[/itex] is not the same as the direction of [itex]\frac{\Delta \vec{v}}{\Delta t}[/itex].
Is that correct?

Yes, Orodruin. So, there's no contradiction, right?
That means, the direction of [itex]\Delta \vec{v}[/itex] is not the direction of [itex]a_{c}[/itex].

And the we should say the direction of [itex]a_{c}[/itex] is what [itex]\Delta \vec{v}[/itex] approaches as [itex]\Delta t[/itex] approaches zero, rather than the direction of [itex]\Delta \vec{v}[/itex].

As long as ##\Delta t## remains finite, the direction of ##\Delta \vec v## is not orthogonal to ##\vec v##. Acceleration is a derivative of the velocity and thus defined in the limit.

Sorry, Orodruin. I want to know more about "thus defined in the limit". Does that mean what I said?

The direction of [itex]a_{c}[/itex] is defined as the direction of what [itex]\Delta \vec{v}[/itex] approaches as [itex]\Delta t[/itex] approaches [itex]0[/itex], rather than the direction of [itex]\Delta \vec{v}[/itex].

Yes. Or rather if you just talk about direction, the direction of ##\vec a## is the what the direction of ##\Delta \vec v## approaches as ##\Delta t \to 0##.