# Definition and (non)equivalent definitions of limit of f

so I am preparing for my final and I needed some help verifying my definitions and their justifications. So any help/feedback would be appreciated...

Let $$f:(0,1) \to \mathbb{R}$$ be a given function, and the limit $$\lim\limits_{x \to x_0} f(x) = L$$. Then out of the following defnition pick the right definition, and/or equuivalent definition. If the definition is not equivalent provide explanation:

a.) for any $$\epsilon \gt 0$$, there exists $$\delta \gt 0$$ 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 true.

b.) 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 $$|f(x) - L|\leq \epsilon$$.

this definition is false. (I don't quite understand how tho)

c.) 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 false because for any ϵ>0 there exists some δ>0 that is small enough. It can't be any delta.

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

this definition is false. suppose $$f(x) = 0$$ then with $$L = 0$$ the limit does not exist anywhere for every$$x$$ as $$|f(x) - L| = 0$$ $$0$$ is not less then $$0$$.

Thank you.

That one

b.) 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 $$|f(x) - L|\leq \epsilon$$

is not correct, indeed by the condition

$$|x-x_0| \leq \delta$$

we could be allowed to take $$x=x_0$$ but in the definition we are assuming a deleted neighborhood of $$x_0$$ that is $$x\neq x_0$$.

The correct version is

b.2) for any $$\epsilon \gt 0$$, there exists $$\delta \gt 0$$ such that for all $$x \in (0,1)$$ and $$\color{red}{0<}|x-x_0| \leq \delta$$, one has $$|f(x) - L|\leq \epsilon$$

• but, since we are not given $0 \lt$ part of the inequality, then it makes the definition invalid, right? – ISuckAtMathPleaseHELPME Dec 16 '18 at 23:06
• oh okay, you edited after i commented but i got it thanks :). does c and d make sense tho? – ISuckAtMathPleaseHELPME Dec 16 '18 at 23:07
• @ISuckAtMathPleaseHELPME Yes exactly, the first version was a little bit confusing! The key is that we are considering a deleted neighborhood of $x_0$. – gimusi Dec 16 '18 at 23:08

Note that for French b) is the right definition

$$\forall \varepsilon>0,\;\exists \delta>0,\; \forall x\in (0,1),\quad \big[|x-x_0|<\delta \implies |f(x)-L|<\varepsilon\big]$$

For other country (UK perhaps) a) is the right definition

$$\forall \varepsilon>0,\;\exists \delta>0,\; \forall x\in (0,1),\quad \big[0<|x-x_0|<\delta \implies |f(x)-L|<\varepsilon\big]$$

and $$b\implies a$$