Show that the limit is in the set I'm having some trouble proving the following :

Let $B = \{(x,y)\in \mathbb{R}^2 : e^x - sin(y)\leq 0\}$ and let $(x_{k},y_{k})_{k \in \mathbb{N}}$ a sequence of elements in B. We suppose that it exists $(x,y) \in \mathbb{R^2}$ such as $\left \| (x_{k},y_{k}) -(x,y)\right \|_{1}\rightarrow _{k\rightarrow +\infty} 0$. 
1) Show that $x$ is the limit of $(x_{k})_{k \in \mathbb{N}}$. What can you say about $(y_{k})_{k \in \mathbb{N}}$ ?
2) Show that $(x,y)$ is in B

The first question is pretty simple because :

$\left \| (x_{k},y_{k}) -(x,y)\right \|_{1} = \left \| (x_{k}-x,y_{k}-y)\right \|_{1} = \left | x_{k}-x \right | + \left | y_{k}-y \right |$
Then we have $\left | x_{k}-x \right |\rightarrow _{k\rightarrow +\infty} 0$ and $\left | y_{k}-y \right |\rightarrow _{k\rightarrow +\infty} 0$
So $x$ is the limit of $(x_{k})_{k \in \mathbb{N}}$ and y is the limit of $(y_{k})_{k \in \mathbb{N}}$

But I have no idea how to prove the second question. I would appreciate some hints. Thanks !

You probably know that exponential and sine are continuous functions. Hence we have $e^{x_k}\to e^x$ and $\sin(y_k)\to \sin(y)$ when $k\to\infty$. Also, since the sequence $(x_k,y_k)$ contains elements of $B$ we have $e^{x_k}-\sin(y_k)\leq 0$ for all $k\in\mathbb{N}$. By taking $k\to\infty$ we get $e^x-\sin(y)\leq 0$. 
