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I would like to prove the following statement:

Consider a function $f:\mathbb{R}\to\mathbb{R}$. Prove that $f$ is continuous at $a\in\mathbb{R}$ if and only if for every sequence $\lbrace a_n\rbrace$ with $\lim_{n\to\infty}a_n=a$, we have $\lim_{n\to\infty}f(a_n)=f(a)$.

Here is my attempt: Suppose that $f(x)$ is continuous at $a\in\mathbb{R}$ and let $\lbrace a_n\rbrace$ be a sequence with $\lim_{n\to\infty}a_n=a$. Because $f(x)$ is continuous, $\lim_{x\to a}f(x)=f(a)$. Let $\epsilon>0$ be arbitrary. Then, there exists $\delta>0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$. By the definition of convergence of a sequence, for every $\epsilon>0$, there exists a number $N\in\mathbb{N}$ such that $|a_n-a|<\epsilon$ for $n>N$. In particular, we can find $N_0\in\mathbb{N}$ such that $n>N_0$ implies $0<|a_n-a|<\delta$. Thus, $|f(a_n)-f(a)|<\epsilon$, which proves $\implies$.

I'm not sure how to approach the proof of the converse direction.

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What you did is correct.

If $f$ is not continuous at $a$, then there is a number $\varepsilon>0$ such, for every $\delta>0$, there is a number $x\in D_f$ such that $\lvert x-a\rvert<\delta$ and $\bigl\lvert f(x)-f(a)\bigr\rvert\geqslant\varepsilon$. In particular, for each $n\in\mathbb N$, there a $x_n\in D_f$ such that $\lvert x_n-a\rvert<\frac1n$ and $\bigl\lvert f(x_n)-f(a)\bigr\rvert\geqslant\varepsilon$. So, $\lim_{n\to\infty}x_n=a$, but $\lim_{n\to\infty}f(x_n)$ either doesn't exist or, if it exists, it is not $f(a)$.

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