To solve the limit [tex]\(\lim_{n \to \infty}\left(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \ldots + \frac{n}{n^2}\right)\)[/tex], let's first rewrite the expression in a more concise form. The limit can be expressed as:
[tex]\[
\lim_{n \to \infty} \sum_{i=1}^{n} \frac{i}{n^2}
\][/tex]
We can factor out [tex]\( \frac{1}{n^2} \)[/tex] from the sum:
[tex]\[
\lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^{n} i
\][/tex]
Next, we need to find the sum of the first [tex]\( n \)[/tex] natural numbers. The formula for the sum of the first [tex]\( n \)[/tex] natural numbers is:
[tex]\[
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}
\][/tex]
Substituting this result into our limit expression gives us:
[tex]\[
\lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{n(n+1)}{2}
\][/tex]
Simplify the expression inside the limit:
[tex]\[
\lim_{n \to \infty} \frac{n(n+1)}{2n^2} = \lim_{n \to \infty} \frac{n^2 + n}{2n^2} = \lim_{n \to \infty} \frac{n^2}{2n^2} + \lim_{n \to \infty} \frac{n}{2n^2}
\][/tex]
[tex]\[
= \lim_{n \to \infty} \frac{1}{2} + \lim_{n \to \infty} \frac{1}{2n}
\][/tex]
As [tex]\( n \)[/tex] approaches infinity, [tex]\(\frac{1}{2n} \)[/tex] approaches [tex]\( 0 \)[/tex]:
[tex]\[
\frac{1}{2} + 0 = \frac{1}{2}
\][/tex]
Hence, the limit is:
[tex]\[
\boxed{\frac{1}{2}}
\][/tex]
