Proving a limit of a constant function

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?

• In fact, with $\textit{any}\;\delta > 0$ it works. – Ángel Mario Gallegos Jun 4 '15 at 19:07

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 $|f(x)-f(10)|=|5-5|=0<\varepsilon$.