Setting up the limit of left-endpoint Riemann sums

Since the interval $[0,\pi/2]$ has length $\pi/2$, dividing it into $n$ equal subintervals gives width $\Delta x = \dfrac{\pi/2 - 0}{n} = \dfrac{\pi}{2n}$. The subintervals are $[x_{i-1}, x_i]$ for $i = 1, \dots, n$, where $x_0 = 0$ is the left end of the whole interval.

The left endpoint of the $i$-th subinterval is $x_i = 0 + (i-1)\Delta x = \dfrac{(i-1)\pi}{2n}$. Substituting $f(x) = \cos(x)$ into the Riemann sum formula gives $\int_0^{\pi/2}\cos(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n \cos\!\left(\frac{(i-1)\pi}{2n}\right)\cdot \frac{\pi}{2n}.$ This expression, in terms of $n$ and $i$ only, is exactly the required limit of left-endpoint Riemann sums.

The width of each subinterval is $\Delta x = \dfrac{\pi}{2n}$. As $n \to \infty$, we have $\Delta x \to 0$. Therefore, the width of each subinterval approaches zero. This is essential for the Riemann sum to converge to the exact area under the curve.