How can I prove that such sequence exists? Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x)=x\sin(x)$. Prove that, for all c $\in \mathbb{R}$, there is a sequence $x_n$ $\in \mathbb{R}$ with $\displaystyle \lim_{n \to \infty}x_n = \infty$ and $\displaystyle \lim_{n \to \infty}f(x_n) = c$.
It's a problem of a book I was using... Actually, I didn't understand how such sequence can exist... I thought that, in this case, if $\displaystyle \lim_{n \to \infty}x_n = \infty$, so $\displaystyle \lim_{n \to \infty}f(x_n) = \infty$. Apparently, I was wrong... 
Can someone help?

For example $c=2$, you can choose one $x$ between $x_1=2n\pi-\frac{\pi}{2}$ and $x_2=2n\pi+\frac{\pi}{2}$, so that $f(x)=2$. This is always possible because $f(x_1)<0<2$ and $f(x_2)=2n\pi+\frac{\pi}{2}>2$ for an infinity number of $n$ and $f(x)$ is continuous.
If $c=\infty$, choose $x_n=2n\pi+\frac{\pi}{2}$.
