I know it’s not the responsibility of this forum to tutor me in calculus, but after doing a whole chapter on limits from OpenStax Calculus Volume One, I’m extremely flustered about how to even begin solving this.
How should I be thinking about this problem in a practical sense? Something alone the lines of as x-> k?
In order to answer your question, let me remind you the definition of continuity of a function.
In the case of a function defined above, you know that for $x <k,$ $f(x)=3x+2$ is a linear function and therefore is continuous on that domain. Moreover, you also know that for $x > k,$ $f(x) =2x-3$ and therefore $f$ is continuous at x for all $x >k$. The only case that we have not checked yet is at $x=k$.
For that, we will use the definition above. Remember that in order for the limit to exists, the limit when you approach $k$ from the left must be the same as the limit when you approach it from the right.
$$\lim_{x \to k^-} f(x) = \lim_{x \to k^-} 3x+2 = 3k+2$$ $$\lim_{x \to k^+} f(x) = \lim_{x \to k^+} 2x-3 = 2k-3.$$
In particular, you need to find a value of $k$ for which these two limits agree. Solving the equation $2k-3 = 3k+2$ gives you that $k =-5.$ There you have $\displaystyle{\lim_{x \to -5^-} f(x) = -13 = \lim_{x \to -5^+} f(x)}.$ Then we can check the second condition, namely $f(-5)= 2(-5)-3 =-13 = \displaystyle{\lim_{x \to -5} f(x)}$, which shows continuity.