Vrftsjtryk

Suppose when $xrightarrow 0$,$f(x) sim x$, $x_n=sum_{i=1}^nf(frac{2i-1}{n^2}a)$
Want to prove: $lim_{n to infty}{x_n}=a(a>0).$

Also,
I don't understand what $sim$ mean?

$f(x)sim x$ means that $f(x)=x+o(x)$ where $o(.)$ is little-o notation. Therefore by substituting$$x_n=a+sum_{i=1}^{n} g(a{2i-1over n^2})$$where $$lim_{xto 0}{g(x)over x}=0$$then you can write $$|g(a{2i-1over n^2})|le a{2i-1over n^2}epsilon$$for large enough $n$ which means that $$left|sum_{i=1}^{n} g(a{2i-1over n^2})right|le sum_{i=1}^{n} |g(a{2i-1over n^2})|le sum a{2i-1over n^2}epsilon=aepsilon$$therefore$$lim_{nto infty}sum g(a{2i-1over n^2})=0$$and $$lim x_n=a$$

Hint: Suppose $f(x)=x$, then we see that
begin{align}
x_n =& frac{a}{n^2}sum^n_{i=1}(2i-1)=frac{2a}{n^2}sum^n_{i=1}i-frac{a}{n}\
=& frac{2a}{n^2}frac{n(n+1)}{2}-frac{a}{n}rightarrow a
end{align}
as $nrightarrow infty$.

Also, $f(x)sim x$ as $x rightarrow 0$ means
begin{align}
lim_{xrightarrow 0}frac{f(x)}{x}=1
end{align}
or more usefully
begin{align}
f(x)=x+text{ something small and goes to 0}
end{align}
when $x$ is close to $0$.