# Let f:D -> R be a continuous function. Let {x_{n}} be a sequence in D that converges to x in D. Then f(X_{n}) -> f(x)

#### G-X

##### New member
Let $$\displaystyle f$$ : $$\displaystyle D$$ -> $$\displaystyle R$$ be a continuous function, where $$\displaystyle D \subset R^{k}$$. Let $$\displaystyle \{x_{n}\}_{n = d}^{\infty}$$ be a sequence in $$\displaystyle D$$ that converges to $$\displaystyle x \in D$$. Then $$\displaystyle f(X_{n})$$ -> $$\displaystyle f(x)$$

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 $$\displaystyle |f(x) - f(x_{n})| = |f(z) - f(x)| < \epsilon$$.

#### G-X

##### New member
We need to show that $$\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)$$.

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 $$\displaystyle |x - x_{n}| = r < \epsilon_{1}$$. Since

$$\displaystyle \delta > 0, ((B_{\delta}(x) \cap D) / \{x\}) \neq \varnothing$$. We know that there $$\displaystyle \exists \epsilon_{1} = \delta$$ meaning that

$$\displaystyle |x - x_{n}| = r < \delta$$, so $$\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D$$.

Let $$\displaystyle a = x \in D$$, thus $$\displaystyle f$$ is continuous at $$\displaystyle x$$, by definition,

$$\displaystyle \forall z \in B_{\delta}(x) \cap D$$ s.t. $$\displaystyle |f(z) - f(x)| < \epsilon_{2}$$. We know that $$\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D$$,

so we can conclude that $$\displaystyle |f(x_{n}) - f(x)| < \epsilon{2}$$.

Thus we have proven that $$\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)$$.

#### G-X

##### New member
We need to show that $$\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)$$.

By definition of convergence, $$\displaystyle |x - x_{n}| < \epsilon_{1}$$. Since $$\displaystyle \delta > 0$$, this implies that $$\displaystyle ((B_{\delta}(x) \cap D) / \{x\}) \neq \varnothing$$.

We know that by definition of convergence this statement holds $$\displaystyle \forall \epsilon_{1} > 0$$, so there $$\displaystyle \exists \epsilon_{1} = \delta$$ implying that $$\displaystyle |x - x_{n}| < \delta$$.

If the $$\displaystyle d(x, x_{n}) < \delta$$ then $$\displaystyle x_{n} \in B_{\delta}(x)$$. We also know that $$\displaystyle x_{n} \in D$$, thus we can conclude that $$\displaystyle x_{n} \in B_{\delta}(x) \cap D$$.

Let $$\displaystyle a = x \in D$$, thus $$\displaystyle f$$ is continuous at $$\displaystyle x$$, by definition, $$\displaystyle \forall z \in B_{\delta}(x) \cap D$$ s.t. $$\displaystyle |f(z) - f(x)| < \epsilon_{2}$$.

We know that $$\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D$$, so we can conclude that $$\displaystyle |f(x_{n}) - f(x)| < \epsilon{2}$$, $$\displaystyle \forall \epsilon_{2} > 0$$ by definition of continuous.

Thus we have proven that $$\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)$$.

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