$\lim\limits_{n \to \infty} \dfrac{1}{\sqrt{n}}\left[1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} + \cdots + \dfrac{1}{\sqrt{n}}\right]$

The given limit is $\lim_{n \to \infty} \dfrac{1}{\sqrt{n}} \left[ 1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} + \dots + \dfrac{1}{\sqrt{n}} \right]$.

This can be written in summation notation as $\lim_{n \to \infty} \dfrac{1}{\sqrt{n}} \sum_{r=1}^{n} \dfrac{1}{\sqrt{r}}$.

To convert this into a Riemann sum, multiplying and dividing the expression by $\sqrt{n}$ gives $\lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{n} \dfrac{\sqrt{n}}{\sqrt{r}}$.

This simplifies to $\lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{n} \dfrac{1}{\sqrt{\dfrac{r}{n}}}$.

Converting the summation into a definite integral using the limit of a sum $\lim_{n \to \infty} \dfrac{1}{n} \sum_{r=1}^{n} f\left(\dfrac{r}{n}\right) = \int_{0}^{1} f(x) dx$, we obtain $\int_{0}^{1} \dfrac{1}{\sqrt{x}} dx$.

Evaluating the definite integral gives $\int_{0}^{1} x^{-1/2} dx = \left[ 2x^{1/2} \right]_{0}^{1} = 2(1) - 2(0) = 2$.

Therefore, the limit equals $2$.