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?
 A: We show something a little stronger: the equation $x\sin x=c$ has arbitrarily large solutions. 
This is an immediate consequence of the Intermediate Value Theorem. For we can make $x\sin x$ arbitrarily large positive or negative. Just use $x=2n\pi+\frac{\pi}{2}$ or $x=2n\pi-\frac{\pi}{2}$.
Let $N$ be any integer such that $2N\pi-\frac{\pi}{2}\gt |c|$. Then for any $n\ge N$ there is a number $x_n$ between $2n\pi-\frac{\pi}{2}$ and $2n\pi+\frac{\pi}{2}$ such that $x_n\sin x_n=c$.
Added: We give more detail. Let $n$ be an integer $\ge N$, where $N$ was described above. Let $u_n=2n\pi -\frac{\pi}{2}$. Then $u_n\sin u_n=-u_n\lt |-c|.  
Let $v_n=2\pi n +\frac{\pi}{2}$. Then $v_n\sin v_n \gt |c|$.
By the Intermediate Value Theorem, there is a number $x_n$ between $u_n$ and $v_n$ such that $x_n\sin x_n=c$. We have now found the $n$-th term of our sequence. (For $k\lt N$, let $x_k=0$). 
Note that  for $n\ge N$, the $x_n$ are all distinct. This is because we always have $v_n\lt u_{n+1}$, so the intervals $(u_n,v_n)$ are all distinct. Since $x_n$ is in the interval $(u_n,v_n)$ and $u_n\to\infty$ as $n\to\infty$, it follows that $x_n\to\infty$.
From $n=N$ on, $x_n\sin x_n$ is exactly $c$, so the sequence $(x_n\sin x_n)$ converges to $c$ in a trivial way.
If we don't want $x_n\sin x_n$ to be exactly $c$ for all large enough $n$, we can modify the construction slightly. By the Intermediate Value Theorem, there is an $x_n$ between $u_n$ and $v_n$ such that $x_n\sin x_n=c+\frac{1}{n}$.  Then the sequence $(x_n)$ has the property that (for $n\ge N$) $f(x_n)=c+\frac{1}{n}$, and therefore the sequence $(f(x_n))$ converges to $c$ as desired.
A: 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 $x_n=2n\pi+\frac{\pi}{2}$.
A: $$x_n = 2\pi n + \frac{c}{2\pi n}, \quad n \in \mathbb{N}.$$
A: How about $x_n = n \cdot \pi$?  $f(x_n)$ is always $0$ but the sequence increases without bound.  For $c \ne 0$, can you construct a similar increasing sequence (where each term is close to a multiple of $\pi$) such that $f(x_n)$ approaches $c$?
