Answer:
![(1)/(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/bxmy27ptt3ycfft3mc1wptpcyx19hwsh6h.png)
Explanation:
we are given a limit
and we want to simplify it
notice that, the numerator is in a sequence of sum of natural number
recall that,
![\rm \displaystyle \: 1 + 2 + 3 + \dots {\dots }+ n = (n(n + 1))/(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/wx1qj5nlk1igjbp8xdh7nhrugv9fw84tlk.png)
so substitute:
![\displaystyle \lim_(n\to \infty ) \frac{ (n(n + 1))/(2) }{ {n}^(2) }](https://img.qammunity.org/2022/formulas/mathematics/high-school/7j8eysosrjsxo8ts3h6i8jtj27g2g8at9t.png)
now recall L'Hôpital's rule
![\displaystyle \lim_(x \to c) (f(x))/(g(x)) = \lim_(x \to c) (f'(x))/(g'(x))](https://img.qammunity.org/2022/formulas/mathematics/high-school/szyhmrzhhf630ighn5fhvhudk2hflklwd5.png)
first simplify the complex fraction:
![\displaystyle \lim_(n\to \infty ) (n+ 1)/(2n)](https://img.qammunity.org/2022/formulas/mathematics/high-school/avfyptlgrrm627upvw49vzjrbb4n65twsq.png)
apply L'Hôpital's rule:
![\displaystyle \lim_( n\to \infty ) ( (d)/(dn) n+ 1)/( (d)/(dn) 2n)](https://img.qammunity.org/2022/formulas/mathematics/high-school/mzchc2ti9com8j7qlz9hz6sxg38pq8cmc1.png)
simplify:
![(1)/(2)](https://img.qammunity.org/2022/formulas/mathematics/high-school/bxmy27ptt3ycfft3mc1wptpcyx19hwsh6h.png)
and we are done: