Using the definition, prove that $\lim\limits_{x \to 10} 5 = 5$
Solution:
when I apply the definition, i get this
$0< |x - 10| < \delta \Rightarrow |5 - 5 | < \epsilon \Rightarrow 0 < \epsilon$
$0 < \epsilon \Rightarrow |x - 10| < \epsilon$ ,and $|x - 10| < \delta$
So i can take $\delta = \epsilon$, or less
Than
$\forall \epsilon, \epsilon >0, \exists \delta = \epsilon; \forall x \in D_f: 0< |x - 10| < \delta \Rightarrow |5 - 5 | < \epsilon $
Is it correct?
I think it's correct, but it's somewhat confusingly written. Let's see if we can clean it up:
Given $\varepsilon > 0$, you want to choose $\delta > 0$ so that if $|x-10|<\delta$ then $|f(x)-f(10)|<\varepsilon$. For this in fact any $\delta$ will work, since $f$ doesn't depend on $x$. So given whatever $\delta$ you like, $|x-10|<\delta$ guarantees $|f(x)-f(10)|=|5-5|=0<\varepsilon$.
I encourage you to use human language in your proofs of things like this. It's just confusing when written entirely in symbols.
