If $f:D\to \mathbb{R}$ is continuous and exists $(x_n)\in D$ such as that $x_n\to a\notin D$ and $f(x_n)\to \ell$ then $\lim_{x\to a}f(x)=\ell$? Assertion: If $f:X\setminus\left\{a\right\}\to \mathbb{R}$ is continuous and there exists  a sequence $(x_n):\mathbb{N}\to X\setminus\left\{a\right\}$ such as that $x_n\to a$ and $f(x_n)\to \ell$ prove that $\lim_{x\to a}f(x)=\ell$
I have three questions: 1) Is the assertion correct? If not, please provide counter-examples. In that case can the assertion become correct if we require that $f$ is monotonic, differentiable etc.?
2)Is my proof correct? If not, please pinpoint the problem and give a hint to the right direcition. Personally, what makes me doubt it are the choices of $N$ and $\delta$ since they depend on another
3)If the proof is correct, then is there a way to shorten it?
My Proof:
Let $\epsilon>0$. Since $f(x_n)\to \ell$ 
 \begin{equation}
\exists N_1\in \mathbb{N}:n\ge N_1\Rightarrow \left|f(x_n)-\ell\right|<\frac{\epsilon}{2}\end{equation}
Thus, $\left|f(x_{N_1})-\ell\right|<\frac{\epsilon}{2}$ and by the continuity of $f$ at $x_{N_1}$,
 \begin{equation}
\exists \delta_1>0:\left|x-x_{N_1}\right|<\delta_1\Rightarrow \left|f(x)-f(x_{N_1})\right|<\frac{\epsilon}{2}
\end{equation}
Since $x_n\to a$,
 \begin{equation}
\exists N_2\in \mathbb{N}:n\ge N_2\Rightarrow \left|x_n-a\right|<\delta_1\end{equation}
Thus, $\left|x_{N_2}-a\right|<\delta_1$ and by letting $N=\max\left\{N_1,N_2\right\}$,
 \begin{gather}
0<\left|x-a\right|<\delta_1\Rightarrow \left|x-x_N+x_N-a\right|<\delta_1\Rightarrow \left|x-x_N\right|-\left|x_N-a\right|<\delta_1\\
0<\left|x-a\right|<\delta_1\Rightarrow \left|x-x_N\right|<\delta_1+\left|x_N-a\right|
\end{gather}
 By the continuity of $f$ at $x_N$,
 \begin{equation}
\exists \delta_3>0:0<\left|x-x_N\right|<\delta_3\Rightarrow \left|f(x)-f(x_N)\right|<\frac{\epsilon}{2}
\end{equation}
Thus, letting $\delta=\max\left\{\delta_1+\left|x_N-a\right|,\delta_3\right\}>0$ we have that,
 \begin{gather}
0<\left|x-a\right|<\delta\Rightarrow \left|x-x_N\right|<\delta\Rightarrow \left|f(x)-\ell+\ell-f(x_N)\right|<\frac{\epsilon}{2}\Rightarrow \left|f(x)-\ell\right|-\left|f(x_N)-\ell\right|<\frac{\epsilon}{2}\\
0<\left|x-a\right|<\delta\Rightarrow\left|f(x)-\ell\right|<\left|f(x_N)-\ell\right|+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon
\end{gather}
We conclude that $\lim_{x\to a}f(x)=\ell$
Thank you in advance
EDIT: The proof is false. One of the mistakes is in this part:
"Thus, letting $\delta=\max\left\{\delta_1+\left|x_N-a\right|,\delta_3\right\}>0$ we have that,
 \begin{gather}
0<\left|x-a\right|<\delta{\color{Red} \Rightarrow} \left|x-x_N\right|<\delta{\color{Red} \Rightarrow} \left|f(x)-\ell+\ell-f(x_N)\right|<\frac{\epsilon}{2}\end{gather}"
 A: Another example: Let $f:\mathbb R\setminus\{0\}\to\mathbb R$ be defined by $f(x)=\mathrm{sign}(x)$, $x_n=\frac{1}{n}$.  This example is also monotone and differentiable.  However, it is not uniformly continuous, and a uniformly continuous function will have a limit at $a$.
A: Your assertion is wrong. A counterexample is for instance given by the sign function, $sgn : \mathbb R \rightarrow \mathbb R$. The sign function is continuous on $\mathbb R\backslash \{0\}$, but 
$$
\lim_{n\rightarrow \infty} sgn(1/n) = 1, 
$$ 
and 
$$
\lim_{n\rightarrow \infty} sgn(-1/n) = -1. 
$$
Here $(1/n)$ and $(-1/n)$ are both sequences that converge to zero, but the sequences $(sgn(1/n))$ and $sgn(-1/n)$ are very much different.
The mistake in your proof is that the distance between an arbitrary point $x$ that is close to $a$ and members of the sequence does not become arbitrary small, so you don't have something like for all $\delta_3$ there is an $N_3$ such that
$$
\vert x-x_n\vert≤\delta_3, ~~\text{for } n≥N_3.
$$ 
But your proof would need something like this. 
In our counterexample with the function $sgn$ this more or less means that if the sequence is given by $-(1/n)$ then I know something about $sgn(x)$ for negative $x$, but I can not say anything about the function values for positive $x$.
A: You need to have $f(x_n) \to l$ for all sequences $x_n \to a$, not just one sequence.
For example, let $a=(0,0)$ with $f(x,y) = \frac{x y}{x^2+y^2}$. This is continuous on $\mathbb{R}^2 \setminus \{a\}$, and the sequence $x_n=(\frac{1}{n},0) \to a$, with $f(x_n) \to 0$ (excuse abuse of notation), but $f$ is not continuous at $a$.
A: There are some important examples in complex analysis, say $D$ is the unit disk in $\mathbb C = \mathbb R^2$ (so not in $\mathbb R$ as in this question).  Some examples of functions, analytic and hence continuous in $D$ are studied, where radial limits exist, but not tangential limits.  These will be counterexamples to what you ask in that case.
