Proving continuity with sequences

[Carla1985]

Could someone confirm that I've answered this question right please
\[

Prove\ using\ the\ sequence\ definition\ that\ f(x)=10x^2\ is\ continuous\ at\ x_0=0\\
I\ have:\ take\ any\ sequence\ x_n\ converging\ to\ 0.\ Then\ f(x_n)=10x_n^2\ converges\ to\ f(x_0)=10*0^2=0\ so\ it\ is\ continuous.\\
is\ that\ sufficient\ for\ the\ proof?

\]
Thankyou

 

[Prove It]

Could someone confirm that I've answered this question right please
\[

Prove\ using\ the\ sequence\ definition\ that\ f(x)=10x^2\ is\ continuous\ at\ x_0=0\\
I\ have:\ take\ any\ sequence\ x_n\ converging\ to\ 0.\ Then\ f(x_n)=10x_n^2\ converges\ to\ f(x_0)=10*0^2=0\ so\ it\ is\ continuous.\\
is\ that\ sufficient\ for\ the\ proof?

\]
Thankyou

I'm not sure what you mean by the sequence definition (I have not done Real Analysis in a while) but to prove continuity here I would simply show that [math]\displaystyle \begin{align*} |x - 0 | < \delta \implies \left| 10x^2 - 0 \right| < \epsilon \end{align*}[/math].

Working on the second inequality we find

[math]\displaystyle \begin{align*} \left| 10x^2 - 0 \right| &< \epsilon \\ \left| 10x^2 \right| &< \epsilon \\ 10 |x| ^2 &< \epsilon \\ |x| ^2 &< \frac{\epsilon}{10} \\ |x| &< \sqrt{ \frac{\epsilon}{10} } \\ |x - 0| &< \sqrt{ \frac{\epsilon}{10} } \end{align*}[/math]

So let [math]\displaystyle \begin{align*} \delta = \sqrt{ \frac{\epsilon}{10} } \end{align*}[/math] and reverse the process to complete your proof.

 

[Carla1985]

Our continuity definition has two parts:

A function f : D → R, D ⊂ R is continuous at x ∈ D
if either of the following equivalent conditions holds:
(i) for every ε > 0 there exists δ = δ(ε) > 0 such that for y ∈ D
y−x|<δ implies |f(y)−f(x)|<ε;
(ii) for every sequence (x_n)_n∈N, x_n ∈ D, converging to x ∈ D it follows that (f (x_n ))_n∈N converges to f (x), i.e. lim_n→∞ x_n = x implies lim_n→∞ f(x_n) = f(x)

I think for this question we have to use the second part as I've already done the ones using the first part :)

 

[chisigma]

Our continuity definition has two parts:

A function f : D → R, D ⊂ R is continuous at x ∈ D
if either of the following equivalent conditions holds:
(i) for every ε > 0 there exists δ = δ(ε) > 0 such that for y ∈ D
y−x|<δ implies |f(y)−f(x)|<ε;
(ii) for every sequence (x_n)_n∈N, x_n ∈ D, converging to x ∈ D it follows that (f (x_n ))_n∈N converges to f (x), i.e. lim_n→∞ x_n = x implies lim_n→∞ f(x_n) = f(x)

I think for this question we have to use the second part as I've already done the ones using the first part :)

A well known theorem on sequences extablishes that $\lim_{n \rightarrow \infty} f(x_{n})= f(x_{0})$ if and only if $\lim_{ n \rightarrow \infty} x_{n}=x_{0}$ and f(x) is continuous in $x=x_{0}$...Kind regards

$\chi$ $\sigma$

 

[Fantini]

Could someone confirm that I've answered this question right please
\[

Prove\ using\ the\ sequence\ definition\ that\ f(x)=10x^2\ is\ continuous\ at\ x_0=0\\
I\ have:\ take\ any\ sequence\ x_n\ converging\ to\ 0.\ Then\ f(x_n)=10x_n^2\ converges\ to\ f(x_0)=10*0^2=0\ so\ it\ is\ continuous.\\
is\ that\ sufficient\ for\ the\ proof?

\]
Thankyou

Hello Carla! Let us try. We want to prove that the sequence $(f(x_n))$ converges to $f(0)$ for any sequence $(x_n)$ converging to $0$. Let $\varepsilon >0$. From the convergence of $(x_n)$ we know that there is a $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$ we have that $|x_n - 0| < \varepsilon$.

Using our definitions, we know that $f(x_n) = 10x_n^2$ and $f(0) = 0$, therefore we have that $|f(x_n) - f(0)| = |10 x_n^2|$. We want to conclude that this is less than $\varepsilon$, so it is desirable to have $|x_n^2| < \frac{\varepsilon}{10}$ and $|x_n| < \frac{\sqrt{\varepsilon}}{\sqrt{10}}$.

I feel this is where we use the convergence of the sequence $(x_n)$. Since it converges, we can take a $N \in \mathbb{N}$ such that $|x_n| < \sqrt{ \frac{\varepsilon}{10} }$. This argument works because the convergence is for all $\varepsilon >0$, in particular this one. :)

Putting it all together: using the convergence of the sequence $(x_n)$, take $N \in \mathbb{N}$ such that $|x_n| < \sqrt{ \frac{\varepsilon}{10} }$. It follows that $$|f(x_n) - f(0)| = |10x_n^2| < 10 \cdot \left( \sqrt{ \frac{\varepsilon}{10} } \right) = \varepsilon,$$ therefore the sequence $(f(x_n))$ converges to $f(0)$ and the function $f$ is continuous at $0$. :D

I hope this helps.

Regards. (Wave)

 

[Opalg]

Could someone confirm that I've answered this question right please
\[

Prove\ using\ the\ sequence\ definition\ that\ f(x)=10x^2\ is\ continuous\ at\ x_0=0\\
I\ have:\ take\ any\ sequence\ x_n\ converging\ to\ 0.\ Then\ \boxed{f(x_n)=10x_n^2\ converges\ to\ f(x_0)=10*0^2=0}\ so\ it\ is\ continuous.\\
is\ that\ sufficient\ for\ the\ proof?

\]
Thankyou

Your proof is correct apart from giving some justification for the assertion that I have boxed. If you are allowed to quote theorems about limits of products then you should say that is what you are doing here. Otherwise you will need to use something like Fantini's argument in the previous comment.

 

[solakis1]

A well known theorem on sequences extablishes that $\lim_{n \rightarrow \infty} f(x_{n})= f(x_{0})$ if and only if $\lim_{ n \rightarrow \infty} x_{n}=x_{0}$ and f(x) is continuous in $x=x_{0}$...Kind regards

$\chi$ $\sigma$

Such a theorem does not exist

 

[Fantini]

Yes it does. The conditions that $f$ is continuous, for every convergent sequence $x_n \to x_0$ we have that $f(x_n) \to f(x_0)$ and for all $\varepsilon >0$ exists $\delta >0$ such that $0 < |x -x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon$ are all equivalent in $\mathbb{R}$, or more generally, in metric spaces.