Use the definition of continuity and the formal definition of the limit value ("$\epsilon$-$\delta$ definition") to show the following: Let $f: \mathbb{R} \to \mathbb{R}$ be a function that is continuous in $x = 0$ and has the property that if $x\neq 0$ then $f (x) \ge 0$. Then $f(0) \geq 0$.
I don't know where to start and I'm not good at using the formal definition of the limit value. I really appreciate some help :)
Before I start:
Suppose I told you I had a number $a$. And I told you that for every $\epsilon > 0$ there will a number $d \ge 0$ so that $|d-a| <\epsilon$.
I claim that that means $a \ge 0$.
Pf:
Either $a = 0$ or $a < 0$ or $a > 0$. If $a=0$ we have nothing to prove. So assume $a \ne 0$.
Let $\epsilon = |a| > 0$ and let $d$ be one of those positive $d$. Then $|d-a| < |a|$. If $a < 0$ then $|a|>|d-a| = d + |a| \ge |a|$ which is a contradiction. So for the claim to be true, we must have $a \ge 0$.
Now to your problem:
Because $f$ is continuous at $x=0$ then by definition.
For every $\epsilon > 0$ there exists a $\delta$ so that if $|x-0| = |x| < \delta$ then $|f(x) -f(0)|< \epsilon$.
If $f(0)=0$ we have nothing to prove. If $f(0) \ne 0$ then $|f(0)| > 0$.
Let $\epsilon = |f(0)|$ and $\delta$ the number that makes it so that if $|x| < \delta$ then $|f(x)-f(0)| < |f(0)|$. And let $x: 0 < x < \delta$. ANd we know $f(x) \ge 0$.
Then we have $|f(x) -f(0)| < |f(0)|$. So by my claim at the start of this post: $f(0) \ge 0$.