Can we find the limit $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^nx^2}{n^2+x^2}$ without evaluating the sum? How to find the limit   $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{2}}{n^{2}+x^{2}}$   if we don't evaluate the sum?
I know the sum is actually an elementary function which we can find it using Fourier series or other methods, but I'm just curious about if there exists some alternative ways to find this limit.
I tried to write it as this form: 
$$\displaystyle\lim_{x\rightarrow\infty}x\sum_{n=1}^{\infty}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right).$$ 
As we know, $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{x}{\left(2n\right)^{2}+x^{2}}$ and $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{x}{\left(2n-1\right)^{2}+x^{2}}$ must get a same value (we don't need to care about what the exact value is) , so this is in the form $``0\cdot\infty"$, which cannot be evaluated directly. This is where I get stucked.  
After days of thinking, I'm getting closer to the answer.
We can use easy algebra to get that $$\left |\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right |\leq\frac{1}{n^2}\quad\forall n\in\mathbb{Z^+},x\in\mathbb{R}$$
Hence the series below converges uniformly on $\mathbb{R}$:$$\displaystyle\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)$$
Changing the order of sum and limit, we can get:$$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)=0$$
which is$$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)=\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)$$
and we also know $$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)=\lim_{x\rightarrow\infty}\left(-\frac{x^2}{1+x^2}-\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)\right)$$
If the limit exists, there must be an equation for the limit $L=-1-L$ which solves $L=-1/2$.
So everything needed is to prove that the limit exists. This would require a bit of analysis.
I’m going to prove it via Cauchy’s rule ($\displaystyle\lim_{x\rightarrow+\infty}f\left(x\right)\ exists\Leftrightarrow\forall\epsilon>0\exists X>0 \forall x_1,x_2>X, \left|f(x_1)-f(x_2)\right|<\epsilon$).

Thanks for @Fimpellizieri , but I think I have got the answer.
We know $$\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}=\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}$$
Consider
$$\left|\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right|\leq\frac{1}{n^2},\ \forall x\in\mathbb{R},n\in\mathbb{Z^+}$$
By Weierstrass M test the series below converges uniformly on $\mathbb{R}$. We have $$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left[\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right]\\=\sum_{n=1}^{\infty}\lim_{x\rightarrow\infty}\left[\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right]=0.$$
So we just need to consider $\displaystyle\sum_{n=1}^{\infty}\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}=\frac{1}{x}\sum_{n=1}^{\infty}\frac{\frac{-4n}{x}}{\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$.
Then consider the "Riemann sum" of function $f\left(u\right)=\frac{-4u}{\left(4u^2+1\right)^2}$ on $\left[0,\infty\right)$ , we know the limit of the sum as $x\rightarrow\infty$ is actually the integral $$\int_{0}^{\infty}f\left(u\right)du=\frac{1}{8u^2+2}|_{0}^{\infty}=-\frac{1}{2}$$
which is the final answer.  
Now I explain why the sum tends to the improper integral.
Notice that $f\left(u\right)$ is decreasing on $\left[0,\frac{1}{2}\right]$ and increasing on $\left[\frac{1}{2},+\infty\right)$, we split the sum into two parts $\displaystyle\sum_{0<2n\leq x}\frac{\frac{-4n}{x}}{x\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$ and $\displaystyle\sum_{2n>x}\frac{\frac{-4n}{x}}{x\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$ , noted respectively as $S_{1}\left(x\right)$ and $S_{2}\left(x\right)$.
$S_{1}\left(x\right)$ is obviously tending to the integral of $f\left(u\right)$ on $\left[0,\frac{1}{2}\right]$ , while $S_{2}\left(x\right)$ is bounded between $\displaystyle\int_{\frac{1}{x}\left(\left[\frac{x}{2}\right]+1\right)}^{\infty}f\left(u\right)du$ and $\displaystyle\int_{\frac{1}{x}\left(\left[\frac{x}{2}\right]+2\right)}^{\infty}f\left(u\right)du$ (due to monotonicity) , which both tend to $\displaystyle\int_{\frac{1}{2}}^{\infty}f\left(u\right)du$ .
