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We need to show that \(\displaystyle f(X_{n})\) -> \(\displaystyle f(x)\) or \(\displaystyle lim_{x_{n} -> \infty}f(x_{n}) = f(x)\).

We know that f is continuous at \(\displaystyle a \in D\). We also know that \(\displaystyle f\) is

continuous at \(\displaystyle a\), by definition, \(\displaystyle \forall z \in B_{\delta}(a) \cap D\) s.t. \(\displaystyle |f(z) - f(a)| < \epsilon\).

Since, \(\displaystyle \{x_{n}\}_{n = d}^{\infty}\) is a sequence in \(\displaystyle D\) that converges to \(\displaystyle x \in D\) then there

\(\displaystyle \exists N_{\epsilon}\) such that \(\displaystyle n \ge N_{\epsilon}\) then \(\displaystyle |f(x) - f(x_{n})| < \epsilon\). Let \(\displaystyle a = x \in D\), thus

we can conclude that \(\displaystyle |f(x) - f(x_{n})| = |f(z) - f(x)| < \epsilon\). We now need to

show that \(\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D\). By definition of convergence, we know

\(\displaystyle \forall \epsilon > 0\) that the \(\displaystyle d(x, x_{n}) < \epsilon\). By definition of continuous, we know that

\(\displaystyle B_{\delta}(x) \cap D \neq \varnothing\). By definition of neighborhood, \(\displaystyle N(x, \delta) =

\) \{\(\displaystyle y \in R : d(x, y) < \delta\}\). This implies that \(\displaystyle \exists \epsilon = \delta\) such that at least a single

\(\displaystyle y \in B_{\delta}(x) \cap D\). Since the sequence is bounded by \(\displaystyle x_{n}\) this implies \(\displaystyle \exists z = x_{n} = y\).

We have shown that \(\displaystyle |f(x) - f(x_{n})| = |f(z) - f(x)| < \epsilon\).