# explain why the following definition is not equivalent to the definition of the limit of function f

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 $$\epsilon \gt 0$$ there exists $$\delta \gt 0$$ such that for all $$x \in (0,1)$$ and $$|x-x_0| \leq \delta$$, one has $$|f(x) - L| < \varepsilon$$ .

I don't see the difference between this definition and the actual definition of the limit of a function. Could someone help me out with this?

Please and thank you.

• You have to exclude $x=x_0$, that's why the standard definition has $0<|x-x_0|≤\delta$. If, say, your function is $f(x)=1$ for $x\neq \frac 12$ and $f\left( \frac 12\right) =0$ then $\lim_{x\to \frac 12}f(x)=1$ with the usual definition, but not with your definition. – lulu Dec 11 '18 at 0:56
• Wait so do you mean to say that it’s suppose to be $0 \lt |x- x_0| \lt \delta$ And not less than or equal to delta? – ISuckAtMathPleaseHELPME Dec 11 '18 at 18:40
• That's not a meaningful difference. Think it through...you can always modify $\delta$ according to which definition you favor. But the $>0$ requirement has real meaning. – lulu Dec 11 '18 at 19:05