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Let $f:(0,1) \to \mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit

$\lim\limits_{x \to x_0} f(x) = L$

of $f$ at $x_0 \in [0,1]$ .

For any $\epsilon \gt 0$,for any $\delta \gt 0$ such that for all $x \in (0,1)$ and $0 \lt|x-x_0| \lt \delta$, one has $|f(x) - L| \lt \epsilon$ .

This definition is incorrect because for any $\epsilon \gt 0$ there exists some $\delta \gt 0$ that is small enough. It can't be any delta. Is this the only reason why this definition is not valid?

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    $\begingroup$ Please change your username to something less self-insulting?? $\endgroup$ – Namaste Dec 16 '18 at 19:31
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Yes the correct definition requires

$$\forall \epsilon >0 \quad \exists \delta >0 \quad \ldots$$

and the other part of the definition is correct.

Indeed let consider for example $f(x)=x$ with $\lim_{x\to 1} x=1$ and take $\epsilon =.01$ and $\delta =.5$ then assume $x$ such that $0<|x-1|<0.5$ that is $x=1.4$ and we have

$$|f(x)-1|=|1.4-1|=0.4>\epsilon$$

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  • $\begingroup$ but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0?? $\endgroup$ – ISuckAtMathPleaseHELPME Dec 16 '18 at 19:29
  • $\begingroup$ @ISuckAtMathPleaseHELPME You can consider a simple case for example $f(x)=x$ and try to apply the other definition to see that it doesn't work. $\endgroup$ – gimusi Dec 16 '18 at 19:32
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Your definition implies the known definition but the converse is not true.

Take $f$ defined by :

$$f(x)=0 \;\; \text{ if } \;\; x<\frac 12$$ and $$f(x)=\color{red}{4}\;\; \text{ if } \;\; x\ge \frac 12.$$

then we have $$\lim_{x\to 0^+}f(x)=0$$ but

if we take $\epsilon = \color{red}{3}$

then we do not have $$\forall \eta>0 \;\; \forall x\in(0,1)$$ $$|x-0|<\eta \implies \;\; |f(x)-0|<3$$

for example $f(\frac{9}{10})=\color{red}{4}>\epsilon$

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