Why isn't $\lim\limits_{x\to 3}=6$ true in these conditions? Let $\left(u_n\right)$ be a succession with general term:
$$u_n=3+\frac{(-1)^n}{n}$$ 
and $h$ a real function such that $\lim h\left(u_n\right)=6$. 
This is a multiple choice question and I'm pretty sure the right answer is the following:
If $h$ is continuous in $x=3$ then $h(3)=6$.
But why can't we say that $\lim\limits_{x\to 3}h(x)=6$ (this is another option)?
I'm thinking that if we have $n$ even or odd we could define two successions such that $v_n\to 3^-$ and $w_n\to 3^+$ and that would mean $\lim\limits_{x\to 3}h(x)=6$. Am I wrong?

The point is this:
$\lim_{n\to\infty}h(x_n)=6$ means that the sequence $(h(x_n))_{n\in\mathbb{N}}$ (with the given sequence $(x_n)_{n\in\mathbb{N}}$) converges to 6.
But $\lim_{x\to 3}h(x)=6$ means, that for every sequence $(x_n)_{n\in\mathbb{N}}\subset \mathbb{R}\setminus \{3\}$ that converges to $3$ it holds that the sequence $(h(x_n))_{n\in\mathbb{N}}$ converges to $6$. 
Now obviously, from the first one, you can't follow the secound one.
