Answer:
[tex] \dfrac{1}{2} [/tex]
Step-by-step 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,
[tex] \rm \displaystyle \: 1 + 2 + 3 + \dots {\dots }+ n = \frac{n(n + 1)}{2} [/tex]
so substitute:
[tex]\displaystyle \lim_{n\to \infty } \frac{ \dfrac{n(n + 1)}{2} }{ {n}^{2} } [/tex]
now recall L'Hôpital's rule
[tex] \displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} [/tex]
first simplify the complex fraction:
[tex]\displaystyle \lim_{n\to \infty } \frac{n+ 1}{2n} [/tex]
apply L'Hôpital's rule:
[tex]\displaystyle \lim_{ n\to \infty } \frac{ \dfrac{d}{dn} n+ 1}{ \dfrac{d}{dn} 2n} [/tex]
simplify:
[tex] \dfrac{1}{2} [/tex]
and we are done:
Step-by-step explanation:
solution given:
The given expression takes a form[tex] \frac{ \infty }{ \infty } [/tex]
when n=[tex] \infty [/tex]
so,
[tex] \displaystyle \lim_{n\to \infty } \frac{1 + 2 + 3 + \dots {\dots} +n }{ {n}^{2} } [/tex]
[tex] \displaystyle \lim_{n\to \infty }\frac{n(n+1)}{ 2{n}^{2} }[/tex]
[tex] \displaystyle \lim_{n\to \infty }\frac{n+1}{ 2n}[/tex]
[tex] \frac{1}{2} [/tex]
[tex] \displaystyle \lim_{n\to \infty }[1+\frac{1}{ n}][/tex]
=[tex] \frac{1}{2} [/tex](1+[tex] \frac{1}{ \infty }) [/tex]
=[tex] \frac{1}{2} [/tex] is your answer
