This is a good question! It is all a matter of how to define an improper integral. The traditional way is to write
$$\int_a^b f(t)\, dt = \lim_{x \to a} \lim_{y \to b} \int_x^y f(t)\, dt$$
whenever one of the bounds is worrisome. In particular, we can write
$$\int_a^b f(t) \,dt = \lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \, dt + \int_1^y f(t)\, dt \right)$$
or
$$\lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \,dt \right) + \lim_{x \to a} \lim_{y \to b} \left( \int_1^y f(t) \,dt \right) $$
so that
$$\int_a^b f(t)\, dt = \lim_{x \to a} \left(\int_x^1 f(t) \,dt \right) + \lim_{y \to b} \left(\int_1^y f(t)\, dt \right).$$
Now convergence means that the integral is finite and the only way for the sum of two numbers to be finite is if both are finite. Hence your professor's comment. All of the above assumes $a < 1 < b$ and that there is no trouble with $f$ near $t=1$.
But you are right that you can get cancelation if you take the limits differently. Instead of taking the limit to a and then the limit to b sequentially, if you took them at the same time--that is to say you combined them--you could possibly get cancellation.
Here's an example. If define instead the improper integral as
$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \int_{-x}^x \tan t\,dt$$
then
$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \left(- \ln \cos(x) + \ln \cos(-x) \right) = 0.$$
What you are thinking of is a useful concept and it often comes up with the term Principal Value in more advanced calculus courses.
The value of an infinite series is unique (when it exists), despite the Riemann Rearrangement Theorem. If you want to get a different sum, then you have to rearrange the terms, giving you a different sequence of terms (albeit consisting of the same values). Similarly, the value of a(n improper) Henstock–Kurzweil integral is unique result (when it exists), even if you can get a different integral by rearranging the values, giving you a different function.
For a specific example (following the comment that @Hugo made), let $ a _ n $ be $ ( - 1 ) ^ { n \! } / n $ (for a positive integer $ n $), and let $ f ( x ) $ be $ a _ { \lfloor x \rfloor } $ (where $ \lfloor x \rfloor $, also written $ [ x ] $, is the floor of $ x $: round $ x $ down to an integer) for real $ x \geq 1 $. Then $ \int _ { x = 1 } ^ \infty f ( x ) \, \mathrm d x = \sum _ { n = 1 } ^ \infty a _ n = - \! \ln 2 $. Now let the sequence $ b $ be a rearrangement of $ a $ such that $ \sum _ { n = 1 } ^ \infty b _ n = 3 $ (for example). We can similarly get a rearrangement $ g $ of $ f $ defined by $ g ( x ) = b _ { \lfloor x \rfloor } $, so that $ \int _ { x = 1 } ^ \infty g ( x ) \, \mathrm d x = 3 $. But $ b $ is not the same sequence as $ a $, and $ g $ is not the same function as $ f $.
The difference between $ g $ and $ f $ is that certain intervals are swapped around. From a measure-theoretic perspective, this should make no difference, since it's the same values defined on sets with the same measure, which is why the HK integral doesn't work as well as the Lebesgue integral in measure theory. But it doesn't stop the HK integral from being well defined.
Best Answer
The limit is $\pi$ irrespective of what those sequences are. It is a standard fact that $\lim _{t\to \infty}\int_0^{t} \frac {\sin \, x} x \, dx=\pi/2$ from which my claim follows.