I am trying to prove that $\lim_{x \to 0}(x+a)=a $, using the epsilon delta definition of limits. Here is my attempt:
Let $0<\left | x \right |<\delta $ and $\left | x+a \right |<\varepsilon$ for $\varepsilon>0$
By triangle inequality:
$\left | x+a \right |\leq \left | x \right |+\left | a \right |<\delta +\left | a \right |<\varepsilon\\\\\Rightarrow \delta < \varepsilon-\left | a \right |$
But this is not true for $\varepsilon \leq \left | a \right |$, since that would give a $\delta < 0$.
Any help would be appreciated.
You have to show:
if $ \epsilon >0$ is given, then there exists $ \delta >0$ such that
$$|f(x)-a| < \epsilon$$
if $0< |x|< \delta.$
Hint: we have $|f(x)-a|=|x|.$