# Proving that a limit exists given information about converging sequences.

Let $(Z,d), (Y,d')$ be metric spaces and $f: X \subset Z \to Y$ a function. Let $a$ be a limit point of $X$ in $Z$. Define the function

$$\hat{f}_b: X \cup\{a\} \to Y: x \mapsto \begin{cases}{f(x) \quad x \neq a} \\ b \quad\quad x = a \end{cases}$$

where $b\in Y$ is a fixed element. Given the following statements:

(1) For any two sequences $(x_n)_n, (y_n)_n$ in $X -\{a\}$ such that $x_n \to a$ and $y_n \to a$, we have: $(f(x_n))_n, (f(y_n))_n$ are convergent sequences and share the same limit.

(2) $\lim_{x \to a} f(x)$ exists

Prove that $(1) \implies (2)$

My attempt:

Call $b$ the unique limit of the image of all sequences that consist of elements in $X - \{a\}$ and converge to $a$.

Since we have that $\lim_{x \to a} f(x) = b \iff \hat{f}_b$ is continuous in $a$, it suffices to show that $\hat{f}_b$ is continuous in $a$.

To do this, we take an arbitrary sequence $(x_n)_n$ such that $x_n \to a$. We must prove that $\hat{f}_b(x_n) \to \hat{f}_b(a) = b$.

If the sequence $(x_n)_n$ contains only a finite amount of terms that are $a$, the sequence $f(x_n)$ converges to $b$.

However, I have trouble proving that $f(x_n) \to b$ if $(x_n)$ contains the term $a$ an infinite amount of times.

Please, post only solutions that prove that $\hat{f}_b$ is continuous at $a$

Actually, you have to prove that $\hat{f}_b(x_n) \xrightarrow{n\to\infty} b$, since $a$ is not in the domain of $f$.

Let $(x_n)_{n=1}^\infty$ be a sequence in $X \cup \{a\}$ such that $x_n \xrightarrow{n\to\infty} a$.

If $a$ occurs only finitely many times in $(x_n)_{n=1}^\infty$ (let $n_0$ be the last occurrence), then the sequence $(x_n)_{n=n_0+1}^\infty$ is in $X \setminus \{a\}$ so $(f(x_n))_{n=n_0+1}^\infty$ converges to $b$. Then $(f(x_n))_{n=1}^\infty$ also converges to $b$.

Assume that $a$ occurs infinitely many times in $(x_n)_{n=1}^\infty$. Let $(x_{q(n)})_{n=1}^\infty$ be the subsequence of $(x_n)_{n=1}^\infty$ containing all elements not equal to $a$.

Using your assumption, $\left(\hat{f}_b(x_{q(n)})\right)_{n=1}^\infty = (f(x_{q(n)}))_{n=1}^\infty$ converges to $b$.

Let's show that $\left(\hat{f}_b(x_n)\right)_{n=1}^\infty$ converges to $b$. Let $\varepsilon > 0$. Since $f(x_{q(n)}) \xrightarrow{n\to\infty} b$, there exists $n_0 \in \mathbb{N}$ such that $n \ge n_0 \implies d'(f(x_{q(n)}), b) < \varepsilon$.

Then, for $n \ge q(n_0)$ we have that $d'\left(\hat{f}_b(x_{n}), b\right) < \varepsilon$, since $d'\left(\hat{f}_b(x_{n}), b\right)$ is either $d'\left(f(x_{q(n)}), b\right)$ for some $n \ge n_0$, or it is $d'\left(\hat{f}_b(a), b\right) = d'(b,b) = 0$.

We conclude that $\hat{f}_b(x_n) \xrightarrow{n\to\infty} b$.

• Great answer! Why the first line though? What did I say that was wrong? – user370967 Nov 23 '17 at 19:47
• @Math_QED It's about your last line: 'I have trouble proving that $f(x_n)\to b$ if $(x_n)$ contains $a\ldots$'. You should be talking about $\hat{f}_{b}(x_n) \to b$ as $f(x_n)$ can be undefined. – mechanodroid Nov 23 '17 at 20:02
• Btw. I omitted the case when $(x_n)_{n=1}^\infty$ has at most finitely many terms different than $a$. In that case $\hat{f}_b(x_n)$ is eventually equal to $b$, so it trivially converges to $b$. – mechanodroid Nov 23 '17 at 20:04
• I see, that was a typo. You are correct! Thanks a lot! – user370967 Nov 23 '17 at 20:05