If $$f(x)=\lim_{n\to \infty}\dfrac{2}{n^2}\left(\sum_{k=1}^{n} kx\right)\cdot \left(\dfrac{3^{nx}-1}{3^{nx}+1}\right)$$ where $n\in \mathbb{N}$, then find the sum of all solutions of the equation $f(x)=|x^2-2|$
Answer: $0$
My attempt:
The first (obvious) thing to be done is finding $f(x)$ and I set about my quest as follows.
We know that $$\lim_{x\to a}h(x)\cdot g(x)=\left(\lim_{x\to a}h(x)\right)\cdot\left(\lim_{x\to a}g(x)\right)$$ when both limits (of $h$ and $g$) exist.
In my case, I let $g(x)=\frac{3^{nx}-1}{3^{nx}+1}$ and the remaining as $h(x)$. It is easy to see that $\lim_\limits{n\to\infty}g(x)=1$ and $\lim_\limits{n\to\infty}h(x)=\lim_\limits{n\to\infty}\frac{2}{n^2}\cdot x \cdot\frac{(n)(n+1)}{2}=x$
Hence, $f(x)=x$ and when you solve it with $|x^2-2|$ the two solutions are $x=1,2$ whose sum of solutions is three instead of the required zero.
Please help. Thanks!
Edit: I now have realized that $$g(x)=\begin{cases} 1,x>0\\-1,x<0\end{cases}$$ but then arrives the problem when $x=0$.
$\lim \frac {3^{nx}-1} {3^{nx}+1}=1$ if $x >0$ and it is $-1$ if $x<0$. [$0$ if $x=0$]. So $f(x)=|x|$ and if a real number $x$ is such that $|x|=|x^{2}-2|$ then $-x$ also satisfies this equation. Hence the sum of the roots is $0$.
[ The roots are $1,-1,2,-2$].