# Uniqueness of limit of sequence and function

I was doing both proof of uniqueness of limit of a sequence and function and i don't understand why in sequence we take $$\nu=max\{\nu',\nu''\}$$ and in function we take $$\delta=min\{\delta',\delta''\}$$

I hope it's clear without writing the whole proof of both theorems, thanks.

Let X,Y be metric spaces, $$f:A\to Y$$ a function and $$l',l''\in Y$$

if $$\lim_{x\to x_0}f(x)=l'$$ and $$\lim_{x\to x_0}f(x)=l'$$ $$\implies l'=l''$$

Proof.

$$\forall\varepsilon>0$$ $$\exists \delta'>0:\forall x\in A$$ $$d_X(x,x_0)<\delta$$ $$x\neq x_0$$ $$d_Y(f(x),l')<\varepsilon$$

$$\forall\varepsilon>0$$ $$\exists \delta''>0:\forall x\in A$$ $$d_X(x,x_0)<\delta$$ $$x\neq x_0$$ $$d_Y(f(x),l'')<\varepsilon$$

Let $$\delta=min\{\delta',\delta''\}$$. So

$$\forall\varepsilon>0$$ $$\exists \delta>0:\forall x\in A$$ $$d_X(x,x_0)<\delta$$ $$x\neq x_0$$ $$d_Y(f(x),l')<\varepsilon$$ $$d_Y(f(x),l'')<\varepsilon$$

Then $$\forall>0$$ $$d_Y(l',l'')<2\varepsilon$$ then $$l'=l''$$

Yes, it is clear. When dealing with sequences, we have a condition of the type$$(\exists\nu\in\mathbb{N}):n\geqslant\nu\implies C_n$$So, if we have a $$\nu_1$$ and a $$\nu_2$$ such that $$n\implies C_n$$ whenever $$n\geqslant\nu_1$$ and $$n\geqslant\nu_2$$, then we take $$\nu=\max\{\nu_1,\nu_2\}$$ because$$n\geqslant\max\{\nu_1,\nu_2\}\iff n\geqslant\nu_1\text{ and }n\geqslant\nu_2.$$But in the case of functions, we have a condition of the type$$\lvert x-a\rvert<\delta\implies\cdots$$Before we had $$\geqslant$$ and now we have $$<$$. So, we consider $$\min\{\delta_1,\delta_2\}$$, because$$\lvert x-a\rvert<\min\{\delta_1,\delta_2\}\iff\lvert x-a\rvert<\delta_1\text{ and }\lvert x-a\rvert<\delta_2.$$