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?
Best Answer
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 $0<\sin(1/t^2)<1$.