Negation of continuity applied to a sequence 
Show that if it is not true that $\lim_{x \to a} f(x)=l$ then $\exists$ $\epsilon$>0 and a sequence $(x_{n}) \rightarrow a$ as $n \rightarrow \infty$ such that $|f(x_{n})-l| \geq \epsilon$.  

Now my attempt at the answer is as follows,
if it is not true that $\lim_{x \to a} f(x)=l$ then
$\exists \epsilon>0$ such that $0<|x-a|<\delta$ implies $|f(x_{n})-l| \geq \epsilon$
so if we define a sequence $(x_{n})$ on $(a-\delta,a+\delta)$ then $\exists$ $\epsilon$>0 and a sequence $(x_{n}) \rightarrow a$ as $n \rightarrow \infty$ such that $|f(x_{n})-l| \geq \epsilon$.    
I am quite certain this isnt a correct proof, or maybe not the full proof, any help on improving/changing the answer would be appreciated.
 A: Nitpicks: 
The negation must be cleaned up in the following manner. "There exists $\epsilon \gt 0$ such that for any $\delta \gt 0$ there is a real number $x $ such that $0\lt |x - a| \lt \delta$ but $|f(x) - l| \ge \epsilon$ ". And when you figure this out you should be able to define a sequence which tends to $a$ and fulfills your condition. For an example define $(x_n)$ by the corresponding contentious value as described above. That is $x_n$ is the troublesome value when $\delta = \frac{\delta}{n}$. Then since $x_n \in (a - \frac{\delta}{n}, a + \frac{\delta}{n})$ for each $n \in \Bbb N$,  $\lim (x_n) = a$ and $|f(x_n) - l| \ge \epsilon$. Hope I helped. 
A: The negation reads as follows: 

There exists $\varepsilon_0>0$ such that, for every $\delta>0$, there are points $x$ in the domain with the property that $0<|x-x_0|<\delta$ and yet $|f(x)-l| \geq \varepsilon_0$.

Fix $\varepsilon_0$ and choose $\delta=\frac{1}{n}$ with $n \in \mathbb{N}$. This yields a sequence $\{x_n\}_n$ as you need.
A: Be carefully with negations! We know that $$
  \lim_{x \to a} f(x) = l \Leftrightarrow \forall \epsilon > 0 \,:\, \exists \delta \,:\, \forall x \,:\, |x-a| < \delta \Rightarrow |f(x) - l| < \epsilon \text{.}
$$
Let's negate that now, using that $\lnot \exists x\, \phi = \forall x\, \lnot \phi$ and that $\lnot \forall x\, \theta = \exists x \, \lnot \theta$. We get $$\begin{eqnarray}
  \lim_{x \to a} f(x) \neq l &\Leftrightarrow&
  \lnot\left(\forall \epsilon > 0 \,:\, \exists \delta >0 \,:\, \forall x, |x-a| < \delta \,:\, |f(x) - l| < \epsilon\right) \\
  &\Leftrightarrow&
  \exists \epsilon > 0 \,:\, \lnot\left(\exists \delta > 0 \,:\, \forall x, |x-a| < \delta \,:\, |f(x) - l| < \epsilon\right) \\
  &\Leftrightarrow&
  \exists \epsilon > 0 \,:\, \forall \delta > 0 \,:\, \lnot\left( \forall x, |x-a| < \delta \,:\, |f(x) - l| < \epsilon\right) \\
  &\Leftrightarrow&
  \exists \epsilon > 0 \,:\, \forall \delta > 0 \,:\, \exists x,|x-a| < \delta \,:\,  |f(x) - l| > \epsilon \text{.} \\
\end{eqnarray}$$
In plain english, $\lim_{x \to a} f(x) \neq l$ thus means that we may pick some $\epsilon > 0$, and then find values for $x$ arbitrarily close to $a$ (i.e., closer than $\delta$ for any $\delta$) for which nevertheless $|f(x) - f(l)| > \epsilon$.
If you read that negation clause by clause, it tells us exactly what we need. It tells us there is some $\epsilon$, such that for every $\delta > 0$ (in particular for $\delta = \frac{1}{n}$) there is an $x$ (call it $x_n$) within $(a-\delta,a+\delta)$ with $|f(x) - l| > \epsilon$. Thus, by collecting all these $x_n$ into a sequence $(x_n$), we found a sequence $(x_n)$ such that $$
  |x_n - a| < \frac{1}{n} \text{ (i.e., our $\delta = \frac{1}{n}$)} \quad \text{and} \quad |f(x_n) - l| > \epsilon \text{.}
$$
Since $|x_n - a| < \frac{1}{n}$ obviously means $x_n \to a$ as $n \to \infty$, we're done.
