Prove that, if $f$ is a non-negative continuous function with domain $[1,\infty$) such that the limit
$$\lim_{n\to\infty}\int_{1}^{n}f(x)dx$$ (where $n$ is an integer) exists, then the improper integral
$$\int_{1}^{\infty}f(x)dx$$
converges and
$$\lim_{n\to\infty}\int_{1}^{n}f(x)dx = \int_{1}^{\infty}f(x)dx$$
I honestly have no idea where to even start with this proof, because to me this looks like the definition of an improper integral. Any thoughts?
You're overlooking an important thing. The initial assumption is the existence of $\lim_{n\to\infty}\int_{1}^{n}f(x)\,dx$, where the limit is taken over integer points $n$ (consider this as the usual limit of a sequence).
You cannot foresee a priori the behaviour of $\int_1^t f(x) \,dx$ when $t$ is not an integer.
That's when the non-negativity of $f$ comes into play: it tells you that the function $t\to \int_1^t f(x) \,dx$ is increasing, and it therefore has a limit as $t$ goes to $\infty$, say $L\in [0,\infty]$.
But it's possible to prove formally (using the very definition of limits) that $$\lim_{n\to\infty, n\in \mathbb N}\int_{1}^{n}f(x)\,dx =\lim_{t\to \infty} \int_1^t f(x) \, dx= L $$
Hence $L$ is finite.