Is it inuitively correct to say that $\lim_{n \to \infty} \frac{x}{n} = \mathrm{d}x$? We know that from the Riemann sum that $\Delta x = \mathrm{d}x$ where $n \to \infty$ and $\Delta x = \dfrac{b - a}{n}$. If, however, $x$ represents some length of interval, can we also say that $\displaystyle\lim_{n \to \infty}\dfrac{x}{n} = \mathrm{d}x$?

I think this is quite a tricky topic to talk about because of how people interpret things vs how they are actually defined. I find it best to think of $dx$ as a separate value, rather than a function or scaling of the variable $x$. whilst we are talking about a small change in $x$ the key word here is change, so rather than $x/n\to0$ really we want the change in $x$, or the interval that we are looking at to be very small rather than the initial or end values.
One nice way to think of this is by looking at your average integral:
$$I_{a,b}[f]=\int_a^bf(x)dx$$
now think of the integral symbol as a stretched out $S$, which is the first letter of sum. Now inside the integral (the part we are summing) is $f(x)\times dx$. In other words we can think of an integral as a sum over our domain $[a.b]$ where the height of each one is $f(x)$ and the width is $dx$. However for our error to tend to zero we require the rectangles to fit the function $f(x)$ well, and actually tend to it exactly. The best way for us to do this is to reduce the width $dx$ to a very small value close to zero, but it cannot actually equal zero or be "infinitely small". Whilst this may seem strange it does not actually matter as we do not need a numerical method for $dx$, unless using some methods for approximating and in that case the statement is not strictly true. Hope this helps
