Finding Limits 3

If $f(x)$ is continuous in $[a,b]$ then the following is true:

$\color{#D61F06}{\lim_{n\to\infty}\frac{b-a}{n}\sum_{k=1}^{n}f(a+\frac{k(b-a)}{n})=\int_{a}^{b}f(x)dx;}$

let $a=0, b=t, f(x)= \sin x$ , then: $\lim_{n\to\infty}\frac{t}{n} \sum_{k=1}^{n}\sin\frac{kt}{n}=\int_{0}^{1} \sin x dx= 1-\cos t,$

and so $\lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^{n-1}\sin \frac{kt}{n}= \frac{1-\cos t}{t},$ using the fact that $\lim_{n\to\infty}\frac{\sin t}{n}=0.$