i try to find limit of equation, where d!=0, $$ \lim_{x\to\infty} \frac{S_n}{a^2_n} $$ i use some formulas to find limit, but it's give me nothing, irreducible fraction: $$ \lim_{x\to\infty} \frac{S_n}{a^2_n} = \frac{\frac{a_1 + a_1 + (n-1)d}{2} * n}{(a_1 + (n-1)d)^2} = \frac{\frac{2a_1n+n^2d-dn}{2}}{(a_1+nd-d)^2} = a_1^2+(nd)^2-d^2 + 2(a_1nd-n(d^2) -a_1d)=a^2_1 + (nd)^2 - d^2+2a_1nd -2n*d^2 -2a_1d =\frac{2a_1n+n^2d-dn}{ 2a^2_1+2(n*d)^2-2d^2-4a_1nd-4n*d^2-4a_1d} $$
where i did mistake and how to find limit, there are should be some easy way, no?
I think you mean $n\rightarrow\infty$ instead of $x\rightarrow\infty$, because the variable does not appear in the limit and everything turns into simple operations with fractions. In solution I'm gonna finding limit if $n\rightarrow\infty$ and $d\neq0$.
$$ \begin{align*} \lim_{n\rightarrow\infty} \frac{S_n}{a_n} &= \lim_{n\rightarrow\infty} \frac{\frac{(a_1 + a_n) n}{2}}{a_n^2} \\ &= \lim_{n\rightarrow\infty} \frac{(a_1 + a_n) n}{2a_n^2} \\ &= \lim_{n\rightarrow\infty} \frac{(2a_1 + d(n-1))n}{2(a_1+d(n-1))^2} \\ &= \lim_{n\rightarrow\infty} \frac{(2a_1 + d(n-1))n}{2(a_1^2+2 a_1 d(n-1) + d^2(n-1)^2)} \\ &= \lim_{n\rightarrow\infty} \frac{(2a_1 + d(n-1))n}{2(a_1^2+2 a_1 d(n-1) + d^2(n-1)^2)} \frac{n^2}{n^2} \\ &= \lim_{n\rightarrow\infty} \frac{\frac{2a_1}{n} + d(1-\frac{1}{n})}{2\left(\frac{a_1^2}{n^2}+2 a_1 d(\frac{1}{n}-\frac{1}{n^2}) + d^2\left(1-\frac{1}{n}\right)^2\right)} \\ &= \frac{d(1-0)}{2(0 + 0 + d^2(1-0))} = \frac{d}{2d^2} = \frac{1}{2d} \end{align*} $$
To be honest, I think it's an obvious answer to this question.