Sorry for posting so many of these! I'm finding them a little tricky and keep running into problems (Angry)Quote:

Prove that $\displaystyle f(x)=\begin{cases} x^2 \ \ \ \ \ \ \ \ \ x<0\\ sin(x) \ \ \ \ x \geq0 \end{cases}$ is continuous at the origin.

I've done two proofs for this, one uses $\displaystyle \epsilon$ and $\displaystyle \delta$, the other uses sequences.

The epsilon-delta version:

I need to find $\displaystyle |f(x)-f(x_0)|<\epsilon$ when $\displaystyle 0<|x-x_0|< \delta$.

Since i'm trying to find continuity at 0, let $\displaystyle x_0=0$.

Therefore I need to find $\displaystyle |f(x)|< \epsilon$ when $\displaystyle 0<|x|< \delta$.

For $\displaystyle x>0$, $\displaystyle |f(x)|< \epsilon \Leftrightarrow \ |sin(x)|< \epsilon \Leftrightarrow \ |x|< \epsilon$. Hence pick $\displaystyle \delta=\epsilon$.

For $\displaystyle x<0$, $\displaystyle |f(x)|<\epsilon \Leftrightarrow \ |x^2|< \epsilon \Leftrightarrow \ |x|< +\sqrt{\epsilon}$. Hence pick $\displaystyle \delta=+\sqrt{\epsilon}$

Therefore choose $\displaystyle \delta=min \{\epsilon, +\sqrt{\epsilon} \}$

$\displaystyle \color{red}{Is \ this \ the \ correct \ way \ to \ write \ \delta?}$

That was preliminary work so I have to write out the proper proof:

Let $\displaystyle \epsilon>0$ and define $\displaystyle \delta:=min \{ \epsilon, +\sqrt{\epsilon} \}$. When $\displaystyle 0<|x|< \delta$:

$\displaystyle |f(x)-f(x_0)|=|x^2|< (\sqrt{\epsilon})^2=\epsilon$ for $\displaystyle x<0$.

$\displaystyle |f(x)-f(x_0)|=|sin(x)| \leq|x|< \epsilon$ for $\displaystyle x>0$.

Hence f is continuous $\displaystyle \forall x \in \mathbb{R}$.

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Now for the limits version.

Let $\displaystyle a_n=-\frac{1}{n}$ and $\displaystyle b_n=\frac{1}{n}$.

Clearly $\displaystyle \lim_{n \rightarrow \infty}a_n=\lim_{a_n \rightarrow 0-} f(a_n)$ and $\displaystyle \lim_{n \rightarrow \infty}b_n=\lim_{b_n \rightarrow 0+} f(b_n)$.

I need to show that $\displaystyle \lim_{a_n \rightarrow 0-}f(a_n)=\lim_{b_n \rightarrow 0+}f(b_n)$ if $\displaystyle \lim_{x \rightarrow 0}f(x)$ exists.

$\displaystyle f(\lim_{n\rightarrow \infty}a_n)=f(\lim_{n \rightarrow \infty}\frac{1}{n})=0$

$\displaystyle f(\lim_{n\rightarrow \infty}b_n)=f(\lim_{n \rightarrow \infty}-\frac{1}{n})=0$

Therefore $\displaystyle \lim_{a_n \rightarrow 0-}f(a_n)=\lim_{b_n \rightarrow 0_+}f(b_n)$ so $\displaystyle \lim_{x \rightarrow 0}f(x)$ exists.

Hence the function is continuous at the origin.

$\displaystyle \color{red}{Is \ this \ correct?}$

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Thanks in advance to anyone who posts! (Rofl)