Improper integral limit 
Hello all, I have two questions today. 
Question 1:
From Calculus 1, I learned that in order for a limit to exist $$ \displaystyle\lim_{x\to a^-} f(x) = \displaystyle\lim_{x\to a^+} f(x) = L $$ In the image that I've attached, you can see improper integration has been carried out; it was then concluded that the limit exists. You can now probably see why I'm uncertain, it has been said that this limit exists. But, the limit was only carried out for $t\to3^-$ and not $t\to3^+$. So is there some kind of exception to this rule when dealing with Improper integrals? 
Question 2:
Before seeing this example, I concluded that an Improper integral was an integral with one of the limits of integration being $\pm\infty$. Clearly, this doesn't agree with the given example. So, I'm tending towards the idea that an improper integral is an integral, in which the function we are integrating is unbounded in the given interval. Would this be correct? 
 A: See limit is considered only at $3^-$ because the function is considered in the domain [0,3], so no need to go to for the right side continuity. This is known as the limit at the boundary points.  Similarly, if you want to find the limit at zero you only have to check the limit at $0^+$.
For your second question, since the denominator goes to infinity when $x$ goes to infinity, that is why it is termed as an improper integral.
The definition given below is taken from Improper Integral.

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration

A: Notice that the original integral does not care what happens outside of $[0,3]$, that is, $\lim\limits_{x\to3^+}$ does not matter.  Particularly, notice that
$$\lim_{b\to3^+}\int_0^b\frac1{\sqrt{3-x}}\ dx=\int_0^3\frac1{\sqrt{3-x}}\ dx+\lim_{b\to3^+}\int_3^b\frac1{\sqrt{3-x}}\ dx$$
As you may notice, taking the limits from the right is actually a bad idea, and indeed, it makes the problem only more improper than what we started with.
As to question 2, an integral is improper if it is not defined for all of the points that you are integrating over.  For example,
$$\int_0^1\sin(1/t^2)\ dt$$
is an improper integral since the integrand is undefined at $t=0$, but it is also bounded, since $0<\sin(1/t^2)<1$.
A: Well, the function $f(x)=\frac{1}{\sqrt{3-x}}$ is Riemann integrable on $[0,3]$. If we define the function $F:[0,3]\to \mathbb{R}$ by
$$F(x)=\int_0^x \frac{1}{\sqrt{3-x}}dx$$ then observe that $F$ is continuous on $[0,3]$. With this,
\begin{align}
\int_0^3\frac{1}{\sqrt{3-x}}dx=F(3)=\lim_{t\to 3^-} F(t)=\lim_{t\to 3^-}\int_0^t\frac{1}{\sqrt{3-x}}dx=\dots
\end{align}
Hope this help.
