help please!!
simplify the following:
[tex] \displaystyle \lim_{n\to \infty } \frac{1 + 2 + 3 + \dots {\dots} +n }{ {n}^{2} } [/tex]
[tex] \text{ note: must provide explanation}[/tex]
​

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

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