Express area under f(x) = 1/x^2 as a limit of right-endpoint Riemann sums

Since $[1,2]$ has length $1$, splitting it into $n$ equal subintervals gives width $\Delta x = \dfrac{2-1}{n} = \dfrac{1}{n}$. The right endpoint of the $i$-th subinterval starts at $a=1$ and adds $i$ widths, so $x_i = 1 + i\Delta x = 1 + \dfrac{i}{n}$.

Substituting $x_i$ into $f(x) = 1/x^2$ gives $f(x_i) = \dfrac{1}{\left(1+\frac{i}{n}\right)^2}$. Multiplying by $\Delta x = \frac{1}{n}$ and summing over $i=1$ to $n$, then taking the limit as $n\to\infty$, produces the desired expression $\int_1^2 \frac{1}{x^2}\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} \frac{1}{\left(1+\frac{i}{n}\right)^2}\cdot\frac{1}{n}.$

Since delta x equals one divided by n, as n grows without bound, delta x approaches zero. This is exactly why the Riemann sum becomes a limit: the subintervals must shrink to zero width to capture the exact area under the curve. The answer is that delta x approaches, or tends to, zero.