Can anyone explain this to me? (Epsilon and Delta)

Question

If I have the statement $\lim_{x\to 0}f(x)=0$ and $$ f(x)=\begin{cases} x,&x>0\\ x-0.3,&x<0 \end{cases} $$ how do I prove that the statement is false?

Is this correct? $[f(x)-0]$ < e for every $[x-0]<d$

So if we choose e= 0.3 then it is $[f(x)]<0.3$ For every $[x]<d$

$-0.3<f(x)<0.3$ for x<d But where do i go next?

Answer 1

Hint.

The definition $\lim_{x\to 0}f(x)=0$ implies that for every $\epsilon>0$, there exists $\delta>0$ such that $-\delta<x<0$ implies $|f(x)-0|<\epsilon$.

But when $x<0$ and close to $0$, $f(x)$ is close to $-0.3$.

So your goal is showing the following:

there exists $\epsilon>0$ such that for every $\delta>0$, there exists $x\in(-\delta,0)$ such that $|f(x)-0|\ge \epsilon$.

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Let $\epsilon=0.1$. For every $\delta>0$, $x=-\frac{\delta}{2}\in(-\delta,0)$ and $$ |f(x)-0|=|x-0.3|=\frac\delta2+0.3>\epsilon. $$ So the statement must be false.

Answer 2

The notation $\lim_{x\to 0^+}f(x)$ means the limit of $f(x)$ as $x$ approaches $0$ through positive values. Precisely, $$L^+=\lim_{x\to 0^+}f(x) \iff$$ $$\iff \forall e>0\;\exists d>0\;\forall x\in (0,d)\;(|f(x)-L^+|<e).$$ Similarly we define $$L^-=\lim_{x\to 0^-}f(x) \iff$$ $$ \iff \forall e>0\;\exists d>0\;\forall x\in (-d,0)\;(|f(x)-L^-|<e).$$ Meanwhile we have $$L=\lim_{x\to 0}f(x) \iff$$ $$\iff \forall e>0\;\exists d>0\;\forall x\in (0,d)\cup (-d,0)\;(|f(x)-L|<e).$$ Show that $$L=\lim_{x\to 0}f(x) \iff \,\lim_{x\to 0^+}f(x)=L=\lim_{x\to 0^-}f(x)\,.$$ For the $f$ of your Q, do $\lim_{x\to 0^+}f(x)$ and $\lim_{x\to 0^-}f(x)\;$ exist, and, if so, what are they?