# Explain why this definition of limit is incorrect

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?

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

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

• but could refresh my memory on why it has to be there exists delta greater than 0 instead of any delta greater than 0?? – ISuckAtMathPleaseHELPME Dec 16 '18 at 19:29
• @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. – gimusi Dec 16 '18 at 19:32

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